Shahhoud is participating in the first Div.3 contest on Codeforces, the first problem was:

Given two strings A and B of equal length N (N is odd), find the minimum number of steps needed to change A into B, or print  - 1 if it's impossible.

In each step, you can choose an odd integer x such that 1 ≤ x

 ≤ N, and reverse the substring of length x that is centered in the middle of A. For example, performing a step with x = 3 on the string "abcde" results in "adcbe" and applying x = 5 on "abcde"results in "edcba".

Can you help Shahhoud solve the problem?

Input

The first line contains one integer T

, the number of test cases.

Each test case consists of two lines, the first contains the string A, and the second contains the string B. (1 ≤ |A| = |B| ≤ 105) |A| = |B| is odd.

Both strings consist of lowercase English letters.

Output

For each test case, print one line containing one integer,  - 1 if A

 can't be changed into B, or the minimum number of steps to change A into B.

Example

Input

1
abcxdef
fecxdba

Output

2

題意

給兩個一樣長的，長度為n(n%2==1)n(n%2==1)奇數的串，每一次都以(n+1)/2(n+1)/2為中心翻轉，每一次翻轉的長度是自己決定的，即每次操作都是翻轉一個x(1<=x<(n+1)/2)x(1<=x<(n+1)/2)為起點，n−x+1n−x+1為終點的串，e.g.

if(x=2),abcde==>adcbeif(x=2),abcde==>adcbe

怎樣用最少的翻轉次數使兩個串相同，輸出最少的次數，如果不能使兩個串相同輸出-11

解

翻轉策略一定從外往裡翻轉最優，複雜度O(N) 翻轉的時候不要真的在字串上操作，不要去swap，只需要一個變數time記錄已經翻了多少次，當time為偶數的時候，當前子串就是原來的樣子，time為奇數的時候當前字串是已經翻轉過一次的 e.g.

設S=abcdefg當time%2==1,i==3S沒有變，但串實際上是S′=abedcfg,我們就按照S′來做當time%2==0,i==3串實際上為S′=abcdefg設S=abcdefg當time%2==1,i==3S沒有變，但串實際上是S′=abedcfg,我們就按照S′來做當time%2==0,i==3串實際上為S′=abcdefg

#include <iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
using namespace std;
int main()
{char s1[100500],s2[100500];
int n,i,t,num,j;
scanf("%d",&t);
while(t--)
{
    cin>>s1>>s2;
    n=strlen(s1);
    for(i=0,j=n-1;i<n;i++,j--)
    {
        if(s1[i]!=s2[i]&&s1[i]!=s2[j])
            break;
    }
    if(i!=n)
        printf("-1\n");
    else
    {
        num=0;
        for(i=0,j=n-1;i<n/2;i++,j--)
        {
            if(num%2==0)
            {
                if(s1[i]!=s2[i])
                    num++;
            }
            else
            {
                if(s1[i]!=s2[j])
                {
                    num++;
                }
            }
        }
        printf("%d\n",num);
    }
}
    return 0;
}