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 java.io.*;
  import java.math.BigInteger;
  
  //code by tishrah_
  public class _practise {
  	
  	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()); 
  		} 
  
  		 int[] ia(int n)
  		{
  			int a[]=new int[n];
  			for(int i=0;i<n;i++)a[i]=nextInt();
  			return a;
  		}
  		 int[][] ia(int n , int m)
  			{
  				int a[][]=new int[n][m];
  				for(int i=0;i<n;i++) for(int j=0;j<m ;j++) a[i][j]=nextInt();
  				return a;
  			}
  		 long[][] la(int n , int m)
  			{
  				long a[][]=new long[n][m];
  				for(int i=0;i<n;i++) for(int j=0;j<m ;j++) a[i][j]=nextLong();
  				return a;
  			}
  		 char[][] ca(int n , int m)
  			{
  				char a[][]=new char[n][m];
  				for(int i=0;i<n;i++)
  					{
  					String x =next();
  					for(int j=0;j<m ;j++) a[i][j]=x.charAt(j);
  					}
  				return a;
  			}
  
  		long[] la(int n)
  		{
  			long a[]=new long[n];
  			for(int i=0;i<n;i++)a[i]=nextLong();
  			return a;
  		}
  
  		String nextLine() 
  		{ 
  			String str = ""; 
  			try
  			{ 
  				str = br.readLine(); 
  			} 
  			catch (IOException e) 
  			{ 
  				e.printStackTrace(); 
  			} 
  			return str; 
  		} 
  	}
  	static void sort(long[] a) 
  	{int n=a.length;Random r=new Random();for (int i=0; i<a.length; i++) {long oi=r.nextInt(n), temp=a[i];a[i]=a[(int)oi];a[(int)oi]=temp;}Arrays.sort(a);}
      static void sort(int[] a) 
      {int n=a.length;Random r=new Random();for (int i=0; i<a.length; i++) {int oi=r.nextInt(n), temp=a[i];a[i]=a[oi];a[oi]=temp;}Arrays.sort(a);}
  	public static long sum(long a[])
  	{long sum=0; for(long i : a) sum+=i; return(sum);}
  	public static long count(long a[] , long x)
  	{long c=0; for(long i : a) if(i==x) c++; return(c);}
  	public static int sum(int a[])
  	{ int sum=0; for(int i : a) sum+=i; return(sum);}
  	public static int count(int a[] ,int x)
  	{int c=0; for(int i : a) if(i==x) c++; return(c);}
  	public static int count(String s ,char ch)
  	{int c=0; char x[] = s.toCharArray(); for(char i : x) if(ch==i) c++; return(c);}
  	public static boolean prime(int n)
  	{for(int i=2 ; i<=Math.sqrt(n) ; i++) if(n%i==0) return false; return true;}
  	public static boolean prime(long n)
  	{for(long i=2 ; i<=Math.sqrt(n) ; i++) if(n%i==0) return false; return true;}
  	 public static int gcd(int n1, int n2)
  	  { if (n2 != 0)return gcd(n2, n1 % n2); else return n1;}
  	 public static long gcd(long n1, long n2)
  	  { if (n2 != 0)return gcd(n2, n1 % n2); else return n1;}
  	 public static int[] freq(int a[], int n)
  	  { int f[]=new int[n+1]; for(int i:a) f[i]++; return f;}
  	 public static int[] pos(int a[], int n)
  	  { int f[]=new int[n+1]; for(int i=0; i<n ;i++) f[a[i]]=i; return f;}
  	 
  	
  	 public static long mindig(long n)
  	 {
  		 long ans=9; while(n>0) { if(n%10<ans) ans=n%10; n/=10; }
  		 return ans;
  	 }
  	 public static long maxdig(long n)
  	 {
  		 long ans=0; while(n>0) { if(n%10>ans) ans=n%10; n/=10; }
  		 return ans;
  	 }
  	  
  	public static boolean palin(String s)
  	{
  		StringBuilder sb = new StringBuilder();
  		sb.append(s);
  		String str=String.valueOf(sb.reverse());
  		if(s.equals(str))
  			return true;
  		else return false;
  	}
  	
  	public static void main(String args[])
  	{
  
  		FastReader in=new FastReader();
  		PrintWriter so = new PrintWriter(new BufferedWriter(new  OutputStreamWriter(System.out)));
  		_practise ob = new _practise();
  		
  	    
  	    int T = in.nextInt();		
  	      //int T = 1;
  		
  	     tc : while(T-->0)
  		 { 	
  	    	 int n = in.nextInt();
  	    	 String s = in.next();
  	    	 int ans=Integer.MAX_VALUE;
  	    	 for(char c='a' ; c<='z' ; c++)
  	    	 {
  	    		 char ch[] = s.toCharArray();
  	    		 int i=0 ,j=n-1;
  	    		 while(i<j)
  	    		 {
  	    			 if(ch[i]!=ch[j])
  	    			 {
  	    				 if(ch[i]==c)
  	    				 {
  	    					 ch[i]=' ';
  	    					 i++;
  	    				 }
  	    				 else if(ch[j]==c)
  	    				 {
  	    					 ch[j]=' ';
  	    					 j--;
  	    				 }
  	    				 else break;
  	    			 }
  	    			 else
  	    			 {
  	    				 i++;
  	    				 j--;
  	    			 }
  	    		 }
  	    		 
  	    		 StringBuilder sb = new StringBuilder();
  	    		 for(char e : ch)
  	    			 if(e!=' ') sb.append(e);
  	    		 String str = String.valueOf(sb);
  	    		 if(palin(str))
  	    		 {
  	    			 
  	    			 ans=Math.min(ans, n-str.length());
  	    		 }
  	    		 
  	    	 }
  	    	 if(ans==Integer.MAX_VALUE) so.println(-1);
  	    	 else so.println(ans);
  		 }
  		
  		
  		so.flush();
  		/*String s = in.next();
  		 * Arrays.stream(f).min().getAsInt()
  		 * BigInteger f = new BigInteger("1"); 1 ke jagah koi bhi value ho skta jo aap 
  		 *                                       initial value banan chahte
  			int a[] = new int[n];
  			 Stack<Integer> stack = new Stack<Integer>();
  			Deque<Integer> q = new LinkedList<>(); or Deque<Integer> q = new ArrayDeque<Integer>();
  			PriorityQueue<Long> pq = new PriorityQueue<Long>();
  			ArrayList<Integer> al = new ArrayList<Integer>(); 
  			StringBuilder sb = new StringBuilder();
  			HashSet<Integer> st = new LinkedHashSet<Integer>();
  			Set<Integer> s = new HashSet<Integer>();
  			Map<Long,Integer> hm = new HashMap<Long, Integer>(); //<key,value>
  			 for(Map.Entry<Integer, Integer> i :hm.entrySet())
  			HashMap<Integer, Integer> hmap = new HashMap<Integer, Integer>();
  			so.println("HELLO");
  			Arrays.sort(a,Comparator.comparingDouble(o->o[0]))
  			 Arrays.sort(a, (aa, bb) -> Integer.compare(aa[1], bb[1]));
  			 Set<String> ts = new TreeSet<>();*/
  	}
  }