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[2001];
    scanf("%d",&n);
    while(n--)
    {
        int ans=0,pp;
        scanf(" %s",&str);
        for(i=0;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=0;
                    for(k=i;k<(i+j+1)/2;k++)
                    {
                        if(str[k]==str[m--])
                            sum++;
                    }
                    if(sum==(j-i+1)/2)
                        ans=max(ans,j-i+1);
                }

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

    }
    return 0;
}