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K. Cyclic Shift(思維)

滴答滴答---題目連結 

K. Cyclic Shift

time limit per test

1.0 s

memory limit per test

256 MB

input

standard input

output

standard output

You are given two strings a and b of the same length and consisting of lowercase English letters. You can pick at most one subsequence of string b

 and do a cyclic shift on that subsequence exactly once.

For example, if you have a string "abcdefg" and you picked the letters at indices 2, 5, and 6 as a subsequence to do a cyclic shift on them, the letter at index 2 will go to index 5, the letter at index 5 will go to index 6, the letter at index 6 will go to index 2, and the string will become "afcdbeg".

Your task is to check if it is possible to make string b equivalent to string a using at most one cyclic shift. Can you?

Input

The first line contains an integer T (1 ≤ T ≤ 200) specifying the number of test cases.

The first line of each test case contains an integer n

 (1 ≤ n ≤ 105) specifying the length of strings a and b. Then two lines follow, giving strings a and b, respectively. Both strings consist only of lowercase English letters.

Output

For each test case, print a single line containing "YES" (without quotes) if it is possible to make string b equivalent to string a using at most one cyclic shift. Otherwise, print "NO" (without quotes).

Example

input

Copy

2
6
abcdef
adcebf
4
abcd
dabd

output

Copy

YES
NO

Note

A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. For example, the sequence  is a subsequence of  obtained after removal of elements , and .

#include<bits/stdc++.h>
using namespace std;
vector<int>q;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        string a,b;
        cin>>a>>b;
        q.clear();
        for(int i=0; i<n; i++)
        {
            if(a[i]!=b[i])
            {
                q.push_back(i);
            }
        }
        int len=q.size();
        bool f=true;
        for(int i=0; i<len; i++)
        {
            if(b[q[i]]!=a[q[(i+1)%len]])
            {
                f=false;
                break;
            }
        }
        if(f)
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}