最长回文子串（Mirrored String II）

Note: this is a harder version of Mirrored string I.

The gorillas have recently discovered that the image on the surface of the water is actually a reflection of themselves. So, the next thing for them to discover is mirrored strings.

A mirrored string is a palindrome string that will not change if you view it on a mirror.

Examples of mirrored strings are "MOM", "IOI" or "HUH". Therefore, mirrored strings must contain only mirrored letters {A, H, I, M, O, T, U, V, W, X, Y} and be a palindrome.

e.g. IWWI, MHHM are mirrored strings, while IWIW, TFC are not.

A palindrome is a string that is read the same forwards and backwards.

Given a string S of length N, help the gorillas by printing the length of the longest mirrored substring that can be made from string S.

A substring is a (possibly empty) string of characters that is contained in another string S. e.g. "Hell" is a substring of "Hello".

Input

The first line of input is T – the number of test cases.

Each test case contains a non-empty string S of maximum length 1000. The string contains only uppercase English letters.

Output

For each test case, output on a line a single integer - the length of the longest mirrored substring that can be made from string S.

Example

Input

Copy

3
IOIKIOOI
ROQ
WOWMAN

Output

Copy

4
1
3
题解：暴力呀。

 #include <iostream>
 #include <stdio.h>
 #include <algorithm>
 #include <string.h>
 #include <math.h>
 #define PI acos(-1.0)
 using namespace std;
 bool judge(char arr)
 {
     if(arr=='A'||arr=='H'||arr=='I'||arr=='M'||arr=='O'||arr=='T'||arr=='U'||arr=='V'||arr=='W'||arr=='X'||arr=='Y')
         return false;
     return true;
 }
 int main()
 {
     int i,j,n,k,m,kk;
     char str[];
     scanf("%d",&n);
     while(n--)
     {
         int ans=,pp;
         scanf(" %s",&str);
         for(i=;i<strlen(str);i++)
             for(j=i;j<strlen(str);j++)
             {
                 if(judge(str[j]))
                     break;
                 if(str[j]==str[i])
                 {
                     m=j;int sum=;
                     for(k=i;k<(i+j+)/;k++)
                     {
                         if(str[k]==str[m--])
                             sum++;
                     }
                     if(sum==(j-i+)/)
                         ans=max(ans,j-i+);
                 }

             }
         printf("%d\n",ans);

     }
     return ;
 }