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.util.*;
  
  import javax.swing.text.Segment;
  
  import java.io.*;
  import java.math.*;
  import java.sql.Array;
  
  
  
  public class Main {
  	  static class Scanner{
  		  
  			BufferedReader br;
  			StringTokenizer st;
  			public Scanner() {
  				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; 
  		    } 
  		}
  	 
  	  static long mod = (long)(1e9 + 7);
  	 
  	static void sort(long[] arr ) {
  		 ArrayList<Long> al = new ArrayList<>();
  		 for(long e:arr) al.add(e);
  		 Collections.sort(al);
  		 for(int i = 0 ; i<al.size(); i++) arr[i] = al.get(i);
  	 }
  	static void sort(int[] arr ) {
  		 ArrayList<Integer> al = new ArrayList<>();
  		 for(int e:arr) al.add(e);
  		 Collections.sort(al);
  		 for(int i = 0 ; i<al.size(); i++) arr[i] = al.get(i);
  	 }
  	static void sort(char[] arr) {
  		ArrayList<Character> al = new ArrayList<Character>();
  		for(char cc:arr) al.add(cc);
  		Collections.sort(al);
  		for(int i = 0 ;i<arr.length ;i++) arr[i] = al.get(i);
  	}
  	static void rvrs(int[] arr) {
  		int i =0 , j = arr.length-1;
  		while(i>=j) {
  			int temp = arr[i];
  			arr[i] = arr[j];
  			arr[j] = temp;
  			i++;
  			j--;
  		}
  	}
  	static void rvrs(long[] arr) {
  		int i =0 , j = arr.length-1;
  		while(i>=j) {
  			long temp = arr[i];
  			arr[i] = arr[j];
  			arr[j] = temp;
  			i++;
  			j--;
  		}
  	}
  
  	static long mod_mul(long a , long b ,long mod) {
  		return ((a%mod)*(b%mod))%mod;
  	}
  	 static long gcd(long a, long b)
  	  {      
  	     if (b == 0)
  	        return a;
  	     return gcd(b, a % b); 
  	  }
  	 static boolean[] prime(int num) {
  			boolean[] bool = new boolean[num];
  		     
  		      for (int i = 0; i< bool.length; i++) {
  		         bool[i] = true;
  		      }
  		      for (int i = 2; i< Math.sqrt(num); i++) {
  		         if(bool[i] == true) {
  		            for(int j = (i*i); j<num; j = j+i) {
  		               bool[j] = false;
  		            }
  		         }
  		      }
  		      if(num >= 0) {
  		    	  bool[0] = false;
  		    	  bool[1] = false;
  		      }
  		      
  		      return bool;
  		}
  
  	 static long modInverse(long a, long m)
  	    {
  	        long g = gcd(a, m);
  	       
  	          return   power(a, m - 2, m);
  	        
  	    }
  	   
  	 static long power(long x, long y, long m){
  	        if (y == 0) return 1; long p = power(x, y / 2, m) % m; p = (int)((p * (long)p) % m);
  	        if (y % 2 == 0) return p; else return (int)((x * (long)p) % m); }
  	   
       static class Combinations{
  	    	 private long[] z;
  	    	 private long[] z1;
  	    	 private long[] z2;
  	    	  public Combinations(long N , long mod) {
  				z = new long[(int)N+1];
  				z1 = new long[(int)N+1];
  				z[0] = 1;
  				for(int i =1 ; i<=N ; i++) z[i] = (z[i-1]*i)%mod;
  			     z2 = new long[(int)N+1];
  				z2[0] = z2[1] = 1;
  			    for (int i = 2; i <= N; i++)
  			        z2[i] = z2[(int)(mod % i)] * (mod - mod / i) % mod;
  			    
  			    
  			    z1[0] = z1[1] = 1;
  			    
  			    for (int i = 2; i <= N; i++)
  			        z1[i] = (z2[i] * z1[i - 1]) % mod;
  			}
  	    	  long fac(long n) {
  	    		  return z[(int)n];
  	    	  }
  	    	  long invrsNum(long n) {
  	    		  return z2[(int)n];
  	    	  }
  	    	  long invrsFac(long n) {
  	    		  return invrsFac((int)n);
  	    	  }
  	    	  long ncr(long N, long R, long mod)
  	    	  {		if(R<0 || R>N ) return 0;
  	    		    long ans = ((z[(int)N] * z1[(int)R])
  	    		              % mod * z1[(int)(N - R)])
  	    		             % mod;
  	    		    return ans;
  	    		}
  	      }
  	      static class DisjointUnionSets {
  			    int[] rank, parent;
  			    int n;
  			  
  			    public DisjointUnionSets(int n)
  			    {
  			        rank = new int[n];
  			        parent = new int[n];
  			        this.n = n;
  			        makeSet();
  			    }
  			  
  			    void makeSet()
  			    {
  			        for (int i = 0; i < n; i++) {
  			          
  			            parent[i] = i;
  			        }
  			    }
  			  
  			    int find(int x)
  			    {
  			        if (parent[x] != x) {
  			        
  			            parent[x] = find(parent[x]);
  			  
  			        }
  			  
  			        return parent[x];
  			    }
  			  
  			    void union(int x, int y)
  			    {
  			        int xRoot = find(x), yRoot = find(y);
  			  
  			        if (xRoot == yRoot)
  			            return;
  			  
  			        if (rank[xRoot] < rank[yRoot])
  			  
  			            parent[xRoot] = yRoot;
  			  
  			        else if (rank[yRoot] < rank[xRoot])
  			  
  			            parent[yRoot] = xRoot;
  			  
  			        else
  			        {
  			            parent[yRoot] = xRoot;
  			  
  			            rank[xRoot] = rank[xRoot] + 1;
  			        }
  			    }
  			}
  	      static int[] KMP(String str) {
  	    	  int n = str.length();
  	    	  int[] kmp = new int[n];
  	    	  for(int i = 1 ; i<n ; i++) {
  	    		  int j = kmp[i-1];
  	    		  while(j>0 && str.charAt(i) != str.charAt(j)) {
  	    			  j = kmp[j-1];
  	    		  }
  	    		  if(str.charAt(i) == str.charAt(j)) j++;
  	    		  kmp[i] = j;
  	    	  }
  	    	  
  	    	  return kmp;
  	      }
  	      
  	      
  /************************************************ Query **************************************************************************************/	  
  	 
  /***************************************** 		Sparse Table	********************************************************/
  	      static class SparseTable{
  	    		
  	    		private long[][] st;
  	    		
  	    		SparseTable(long[] arr){
  	    			int n = arr.length;
  	    			st = new long[n][25];
  	    			log = new int[n+2];
  	    			build_log(n+1);
  	    			build(arr);
  	    		}
  	    		
  	    		private void build(long[] arr) {
  	    			int n = arr.length;
  	    			
  	    			for(int i = n-1 ; i>=0 ; i--) {
  	    				for(int j = 0 ; j<25 ; j++) {
  	    					int r = i + (1<<j)-1;
  	    					if(r>=n) break;
  	    					if(j == 0 ) st[i][j] = arr[i];
  	    					else st[i][j] = Math.max(st[i][j-1] , st[ i + ( 1 << (j-1) ) ][ j-1 ] );
  	    				}
  	    			}
  	    		}
  	    		public long gcd(long a  ,long b) {
  	    			if(a == 0) return b;
  	    			return gcd(b%a , a);
  	    		}
  	    		public long query(int l ,int r) {
  	    			int w = r-l+1;
  	    			int power = log[w];
  	    			return Math.max(st[l][power],st[r - (1<<power) + 1][power]);
  	    		}
  	    		private int[] log;
  	    		void build_log(int n) {
  	    			log[1] = 0;
  	    			for(int i = 2 ; i<=n ; i++) {
  	    				log[i] = 1 + log[i/2];
  	    			}
  	    		}
  	    	}
  	      
  	      
  /********************************************************	Segement Tree	*****************************************************/
  /**
  	 	 static class SegmentTree{
  			 long[] tree;
  			 long[] arr;
  			 int n;
  			 SegmentTree(long[] arr){
  				 this.n = arr.length;
  				 tree = new long[4*n+1];
  				 this.arr = arr;
  				 buildTree(0, n-1, 1);
  			 }
  			 
  			 
  			  void buildTree(int s ,int e  ,int index ) {
  					if(s == e) {
  						tree[index] = arr[s];
  						return;
  					}
  				
  					int mid = (s+e)/2;
  					
  					buildTree( s,  mid, 2*index);
  					buildTree( mid+1, e, 2*index+1);
  					
  					tree[index] = Math.min(tree[2*index] , tree[2*index+1]);
  				}
  			  
  			 long query(int si ,int ei) {
  				 return query(0 ,n-1 , si ,ei , 1   );
  			 }
  			 private long query( int ss ,int se ,int qs , int qe,int index) {
  					
  					if(ss>=qs && se<=qe) return tree[index];
  					
  					if(qe<ss || se<qs) return (long)(1e17);
  					
  					int mid = (ss + se)/2;
  					long left = query( ss , mid , qs ,qe , 2*index);
  					long right= query(mid + 1 , se , qs ,qe , 2*index+1);
  					return Math.min(left, right);
  				}
  			 public void update(int index , int val) {
  				 arr[index] = val;
  				 for(long e:arr) System.out.print(e+" ");
  				 update(index , 0 , n-1 , 1);
  			 }
  			 private void update(int id ,int si , int ei , int index) {
  				 if(id < si || id>ei) return;
  				 if(si == ei ) { 
  					 tree[index] = arr[id];
  					 return;
  				 }
  				 if(si > ei) return;
  				 int mid = (ei + si)/2;
  					
  					update( id,  si, mid , 2*index);
  					update( id , mid+1, ei , 2*index+1);
  					
  					tree[index] = Math.min(tree[2*index] ,tree[2*index+1]);
  			 }
  			  
  		 }
  		 */
  
  /* ***************************************************************************************************************************************************/	 
  	 
  	      static Scanner sc = new Scanner();
  	 static StringBuilder sb = new StringBuilder();
  	 public static void main(String args[]) throws IOException {
  	
  		 int tc = 1;
  		  tc = sc.nextInt();
  		 for(int i = 1 ; i<=tc ; i++) {
  
  //			 sb.append("Case #" + i + ": "  );	// During KickStart && HackerCup
  			 TEST_CASE();
  			 
  		 }
  		 System.out.println(sb);
  	 }
  	 
  	 static void TEST_CASE() {
  		 int n = sc.nextInt();
  		 char[] arr = sc.next().toCharArray();
  		 int ans = -1;
  		 for(int i = 0 ; i<26 ; i++) {
  			 char rc = (char)(i + 'a');
  			 int l = 0 , r = n-1;
  			 int c = 0;
  			 boolean rem = true;
  			 while(l<=r) {
  				 if(arr[l] == arr[r]) {
  					 l++;r--;
  				 continue;}
  				 if(arr[l] == rc) {
  					 l++;
  					 c++;
  					 continue;
  				 }
  				 if(arr[r] == rc) {
  					 r--;
  					 c++;
  					 continue;
  				 }
  				 rem = false;
  				 break;
  			 }
  			 if(rem) {
  				 if(ans == -1 || ans>c) {
  					 ans = c;
  				 }
  			 }
  		 }
  		 sb.append(ans+"\n");
  	 }
  }
  
  	 
  
  /*******************************************************************************************************************************************************/
  
  /**
  
   */
  
   
  
  