description: Grandma Capa has decided to knit a scarf and asked Grandpa Sher to make a pattern for it, a pattern is a string consisting of lowercase English letters. Grandpa Sher wrote a string s of length n.
  
  Grandma Capa wants to knit a beautiful scarf, and in her opinion, a beautiful scarf can only be knit from a string that is a palindrome. She wants to change the pattern written by Grandpa Sher, but to avoid offending him, she will choose one lowercase English letter and erase some (at her choice, possibly none or all) occurrences of that letter in string s.
  
  She also wants to minimize the number of erased symbols from the pattern. Please help her and find the minimum number of symbols she can erase to make string s a palindrome, or tell her that it's impossible. Notice that she can only erase symbols equal to the one letter she chose.
  
  A string is a palindrome if it is the same from the left to the right and from the right to the left. For example, the strings 'kek', 'abacaba', 'r' and 'papicipap' are palindromes, while the strings 'abb' and 'iq' are not.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The next 2 ⋅ t lines contain the description of test cases. The description of each test case consists of two lines.
  
  The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of the string.
  
  The second line of each test case contains the string s consisting of n lowercase English letters.
  
  It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
  
  Output
  
  For each test case print the minimum number of erased symbols required to make the string a palindrome, if it is possible, and -1, if it is impossible.
  
  Example
  
  Input
  
  
  5
  8
  abcaacab
  6
  xyzxyz
  4
  abba
  8
  rprarlap
  10
  khyyhhyhky
  
  
  Output
  
  
  2
  -1
  0
  3
  2
  
  Note
  
  In the first test case, you can choose a letter 'a' and erase its first and last occurrences, you will get a string 'bcaacb', which is a palindrome. You can also choose a letter 'b' and erase all its occurrences, you will get a string 'acaaca', which is a palindrome as well.
  
  In the second test case, it can be shown that it is impossible to choose a letter and erase some of its occurrences to get a palindrome.
  
  In the third test case, you don't have to erase any symbols because the string is already a palindrome.
solution: import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  
  import static java.lang.Math.log;
  import static java.lang.Math.min;
  
  public class Main {
  
      //------------------------------------------CONSTANTS---------------------------------------------------------------
      public static final int MOD = (int) (1e9 + 7);
  
      //---------------------------------------------I/0------------------------------------------------------------------
  
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader()
          {
              br = new BufferedReader(
                      new InputStreamReader(System.in));
          }
  
          String next()
          {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  }
                  catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() { return Integer.parseInt(next()); }
  
          long nextLong() { return Long.parseLong(next()); }
  
          double nextDouble()
          {
              return Double.parseDouble(next());
          }
  
          String nextLine()
          {
              String str = "";
              try {
                  str = br.readLine();
              }
              catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  
      public static void print(String s) {
          System.out.println(s);
      }
  
      public static void main(String[] args) {
          FastReader sc = new FastReader();
  
          int t = 1;
          t = sc.nextInt();
  
          for(int i = 0; i<t; i++) {
              System.out.println(solve(sc));
          }
      }
  
      //------------------------------------------------SOLVE-------------------------------------------------------------
  
      public static long solve(FastReader sc) {
          int n = sc.nextInt();
          String ss = sc.next();
          char[] s = ss.toCharArray();
          if (isPalindrome(ss))
              return 0;
          int res = Integer.MAX_VALUE;
  
          for (int i = 0; i<26; i++) {
              char c = (char) ('a' + i);
              List<Integer> occC = new ArrayList<>();
  
              StringBuilder x = new StringBuilder("");
              int cnt = 0;
              int occs = 0;
              for (int j = 0; j<n; j++) {
                  if (s[j] != c) {
                      x.append(s[j]);
                      occC.add(cnt);
                      occs += cnt;
                      cnt = 0;
                  } else {
                      cnt++;
                  }
              }
              occC.add(cnt);
              occs += cnt;
  
              if (!isPalindrome(x.toString())) {
                  continue;
              }
  
              if (occC.size() == 1) {
                  res = Math.min(res, occs);
                  continue;
              }
  
              if (occC.size() % 2 == 1) {
                  occs -= occC.get(occC.size()/2);
              }
  
              for (int j = 0; j<occC.size()/2; j++) {
                  occs -= 2*Math.min(occC.get(j), occC.get(occC.size()-1-j));
              }
  
              res = Math.min(res, occs);
          }
  
          if (res == Integer.MAX_VALUE) {
              return -1;
          }
  
          return res;
      }
  
      public static boolean isPalindrome(String s) {
          for (int i = 0; i<s.length()/2; i++) {
              if (s.charAt(i) != s.charAt(s.length()-1-i))
                  return false;
          }
          return true;
      }
  
      //-----------------------------------------------FUNCTIONS----------------------------------------------------------
  
      public static String printArray(int[] a) {
          StringBuilder r = new StringBuilder("");
  
          for(int i = 0; i<a.length; i++) {
              r.append(a[i] + " ");
          }
          return r.toString();
      }
  
      public static <T> String printList(List<T> a) {
          StringBuilder r = new StringBuilder("");
  
          for (T x: a) {
              r.append(x.toString() + " ");
          }
          return r.toString();
      }
  
      // first element that is not less than target
      public static int lowerBound(int[] arr, int begin, int end, int target) {
          while(begin < end) {
              int mid = begin + (end - begin) / 2;
              // When the element is less than the mid target
              if(arr[mid] < target)
                  // begin to mid + 1, arr [begin] value is less than or equal target
                  begin = mid + 1;
                  // When the mid target element when greater than or equal
              else if(arr[mid] >= target)
                  end = mid;
          }
          return begin;
      }
  
      // first element that is greater than target
      public static int upperBound(int[] arr, int begin, int end, int target) {
          while(begin < end) {
              int mid = begin + (end - begin) / 2;
              if(arr[mid] <= target)
                  begin = mid + 1;
              else
                  end = mid;
          }
          return begin;
      }
  
      // generate permutations (call search and access through permutations)
      public static class Permutations {
          List<List<Integer>> permutations = new ArrayList<>();
          List<Integer> permutation = new ArrayList<>();
          boolean[] taken;
          int[] a;
  
          public Permutations(int[] a) {
              this.a = a;
              taken = new boolean[a.length];
          }
  
          void search() {
              if (permutation.size()==a.length) {
                  permutations.add(new ArrayList<>(permutation));
              } else {
                  for (int i = 0; i<a.length; i++) {
                      if (taken[i]) {
                          continue;
                      }
                      taken[i] = true;
                      permutation.add(a[i]);
                      search();
                      taken[i] = false;
                      permutation.remove(permutation.size()-1);
                  }
              }
          }
      }
  
      // generate subsets (call search and access subsets)
      public static class Subsets {
          List<List<Integer>> subsets = new ArrayList<>();
          List<Integer> subset = new ArrayList<>();
  
          void search(int[] a, int x) {
              if (x == a.length) {
                  subsets.add(new ArrayList<>(subset));
              } else {
                  subset.add(a[x]);
                  search(a, x+1);
                  subset.remove(subset.size() - 1);
                  search(a, x+1);
              }
          }
      }
  
      public static int gcd(int a, int b) {
          while (b != 0) {
              int t = b;
              b = a % b;
              a = t;
          }
          return a;
      }
  
      public static int lcm(int a, int b) {
          return (a*b)/gcd(a, b);
      }
  
      public static int[] prefixSum(int[] a) {
          int n = a.length;
          int[] ps = new int[n];
          ps[0] = a[0];
  
          for (int i = 1; i<n; i++) {
              ps[i] = ps[i-1] + a[i];
          }
  
          return ps;
      }
  
      public static int[][] prefixSum2D(int[][] a) {
          int n = a.length;
          int m = a[0].length;
          int[][] ps = new int[n][m];
          ps[0][0] = a[0][0];
  
          for (int j = 1; j<m; j++) {
              ps[0][j] = ps[0][j-1] + a[0][j];
          }
          for (int i = 1; i<n; i++) {
              ps[i][0] = ps[i-1][0] + a[i][0];
          }
  
          for (int i = 1; i<n; i++) {
              for (int j = 1; j<m; j++) {
                  ps[i][j] = ps[i-1][j] + ps[i][j-1] - ps[i-1][j-1] + a[i][j];
              }
          }
  
          return ps;
      }
  
      public static double log2(double a) {
          return log(a)/log(2);
      }
  
      //----------------------------------------DATA-STRUCTURES-----------------------------------------------------------
  
      // Range queries (i.e. RMQ) in O(1), others with functions in O(log n)
      public static class SparseTable {
          int[][] st;
          int n;
          int k;
  
          public SparseTable(int[] a) {
              n = a.length;
              k = (int) log2(n);
              st = new int[n][k+1];
  
              for (int i = 0; i<n; i++) {
                  st[i][0] = a[i];
              }
  
              // 1<<j == 2^j
              for (int j = 1; j<=k; j++) {
                  for (int i = 0; i + (1 << j) <= n; i++) {
                      st[i][j] = min(st[i][j-1], st[i + (1 << (j - 1))][j - 1]);
                  }
              }
          }
  
          //not always available O(log n) -- change the constructor to use it
          public long sumQuery(int l, int r) {
              long sum = 0;
              for (int j = k; j>=0; j--) {
                  if ((1 << j) <= r - l + 1) {
                      sum += st[l][j];
                      l += 1 << j;
                  }
              }
  
              return sum;
          }
  
          //always available O(1) -- change this and the constructor for rangeMaxQuery
          public int rangeMinQuery(int l, int r) {
              int j = (int) log2(r - l + 1);
              int min = min(st[l][j], st[r - (1 << j) + 1][j]);
              return min;
          }
  
      }
  
      //Range Sum Dynamic Queries O(log n)
      public static class BinaryIndexedTree {
  
          int[] bit;
  
          public BinaryIndexedTree(int[] a) {
              int n = a.length;
              bit = new int[n+1];
              // Store the actual values in BITree[]
              // using update()
              for(int i = 0; i < n; i++)
                  updateBit(i, a[i]);
          }
  
          // O(log n) update delta at index in the array (do it on the array and then call this function)
          public void updateBit(int index, int delta) {
              index += 1;
  
              while (index <= bit.length) {
                  bit[index] += delta;
  
                  index += index & (-index);
              }
          }
  
          // Computes sum a[0...index] in O(log n)
          public int getSum(int index) {
              int sum = 0;
  
              index += 1;
  
              while (index > 0) {
                  sum += bit[index];
  
                  index -= index & (-index);
              }
  
              return sum;
          }
      }
  
      public static class UnionFind {
  
          int[] link;
          int[] size;
  
          UnionFind(int n) {
              link = new int[n];
              size = new int[n];
              for (int i = 0; i<n; i++) {
                  link[i] = i;
                  size[i] = 1;
              }
          }
  
          int find(int x) {
              while (x != link[x]) x = link[x];
              return x;
          }
  
          boolean same(int a, int b) {
              return find(a) == find(b);
          }
  
          void unite(int a, int b) {
              a = find(a);
              b = find(b);
  
              if (size[a] < size[b]) {
                  int temp = a;
                  a = b;
                  b = temp;
              }
  
              size[a] += size[b];
              link[b] = a;
          }
      }
  
      public static class SegmentTree {
  
          int st[];
  
          public SegmentTree(int arr[]) {
              int n = arr.length;
  
              int x = (int) (Math.ceil(log2(n)));
  
              int maxSize = 2 * (int) Math.pow(2, x) - 1;
  
              st = new int[maxSize];
  
              build(arr, 0, n-1, 0);
          }
  
          int build(int arr[], int ss, int se, int si) {
              if (ss == se) {
                  st[si] = arr[ss];
                  return arr[ss];
              }
  
              int mid = ss + (se - ss) / 2; //ss+se/2
  
              st[si] = build(arr, ss, mid, si * 2 + 1) +
                      build(arr, mid + 1, se, si * 2 + 2);
              return st[si];
          }
  
          private int getSumUtil(int ss, int se, int l, int r, int si) {
              if (l <= ss && r >= se) {
                  return st[si];
              }
  
              if (se < l || ss > r) {
                  return 0;
              }
  
              int mid = ss + (se - ss) / 2;
              return getSumUtil(ss, mid, l, r, 2 * si + 1) +
                      getSumUtil(mid + 1, se, l, r, 2 * si + 2);
          }
  
          void updateValueUtil(int ss, int se, int i, int diff, int si) {
              if (i < ss || i > se) {
                  return;
              }
  
              st[si] = st[si] + diff;
              if (se != ss) {
                  int mid = ss + (se - ss) / 2;
                  updateValueUtil(ss, mid, i, diff, 2 * si + 1);
                  updateValueUtil(mid + 1, se, i, diff, 2 * si + 2);
              }
          }
  
          int getSum(int n, int l, int r) {
              if (l < 0 || r > n - 1 || l > r) {
                  return -1;
              }
  
              return getSumUtil(0, n-1, l, r, 0);
          }
  
          // don't do it directly on the array but use this function that will
          // do everything for you
          void updateValue(int arr[], int n, int i, int newValue) {
              if (i < 0 || i > n - 1) {
                  return;
              }
  
              int diff = newValue - arr[i];
  
              arr[i] = newValue;
  
              updateValueUtil(0, n - 1, i, diff, 0);
          }
      }
  
      // Similar to Pair in c++, non-null objects as parameters
      public static class Pair<T extends Comparable<T>, Q extends Comparable<Q>> implements Comparable<Pair<T, Q>>{
          T a;
          Q b;
  
          public Pair(T a, Q b) {
              this.a = a;
              this.b = b;
          }
  
          @Override
          public boolean equals(Object o) {
              if (this == o) return true;
              if (o == null || getClass() != o.getClass()) return false;
              Pair<?, ?> pair = (Pair<?, ?>) o;
              return a.equals(pair.a) && b.equals(pair.b);
          }
  
          @Override
          public int hashCode() {
              return Objects.hash(a, b);
          }
  
  
          @Override
          public int compareTo(Pair<T, Q> tqPair) {
              int compareA = this.a.compareTo(tqPair.a);
  
              if (compareA == 0) {
                  return this.b.compareTo(tqPair.b);
              } else {
                  return compareA;
              }
          }
      }
  
      // Uses Map + TreeSet to get the structure
      public static class OrderedMultiSet<T extends Comparable<T>> {
          private final TreeSet<T> set = new TreeSet<>();
          private final Map<T, Integer> frequencies = new HashMap<>();
          private int size = 0;
  
          public OrderedMultiSet() {}
  
          public OrderedMultiSet(List<T> a) {
              for (T x: a) {
                  this.add(x);
                  size++;
              }
          }
  
          public void add(T element) {
              set.add(element);
              if (!frequencies.containsKey(element)) {
                  frequencies.put(element, 0);
              }
              frequencies.put(element, frequencies.get(element) + 1);
              size++;
          }
  
          public boolean remove(T element) {
              if (frequencies.containsKey(element)) {
                  int x = frequencies.get(element);
                  x--;
                  if (x == 0) {
                      frequencies.remove(element);
                      set.remove(element);
                  } else {
                      frequencies.put(element, frequencies.get(element) - 1);
                  }
                  size--;
                  return true;
              }
              return false;
          }
  
          public int size() {
              return size;
          }
  
          public T first() {
              return set.first();
          }
  
          public T last() {
              return set.last();
          }
  
          public T pollFirst() {
              T t = set.first();
              remove(t);
              return t;
          }
  
          public T pollLast() {
              T t = set.last();
              set.
              remove(t);
              return t;
          }
      }
  }