Codeforces Round #599 (Div. 2) B1. Character Swap (Easy Version)

Posted 2020-11-29 qingyuyyyyy

This problem is different from the hard version. In this version Ujan makes exactly one exchange. You can hack this problem only if you solve both problems.

After struggling and failing many times, Ujan decided to try to clean up his house again. He decided to get his strings in order first.

Ujan has two distinct strings ss and tt of length nn consisting of only of lowercase English characters. He wants to make them equal. Since Ujan is lazy, he will perform the following operation exactly once: he takes two positions ii and jj (1≤i,j≤n1≤i,j≤n, the values ii and jj can be equal or different), and swaps the characters sisi and tjtj. Can he succeed?

Note that he has to perform this operation exactly once. He has to perform this operation.

Input

The first line contains a single integer kk (1≤k≤101≤k≤10), the number of test cases.

For each of the test cases, the first line contains a single integer nn (2≤n≤1042≤n≤104), the length of the strings ss and tt.

Each of the next two lines contains the strings ss and tt, each having length exactly nn. The strings consist only of lowercase English letters. It is guaranteed that strings are different.

Output

For each test case, output "Yes" if Ujan can make the two strings equal and "No" otherwise.

You can print each letter in any case (upper or lower).

Example

input

Copy

4
5
souse
houhe
3
cat
dog
2
aa
az
3
abc
bca

output

Copy

Yes
No
No
No

Note

In the first test case, Ujan can swap characters s1s1 and t4t4, obtaining the word "house".

In the second test case, it is not possible to make the strings equal using exactly one swap of sisi and tjtj.

 

#include<bis/stdc++.h>
using namespace std;
int main() {
    int t;
    scanf("%d",&t);
    while(t--) {
        int n;
        scanf("%d",&n);
        string s,t;
        cin>>s;
        cin>>t;
        int c1=-1,c2=-1;
        int flag = 0;
        int sum=0;
        for(int i = 0; i < n; i++) {
            if(s[i] != t[i]) {
                sum++;
                if(sum == 1) {  //记录不同的位置
                    c1 = i;
                } else if(sum == 2) {
                    c2 = i;
                } else {
                    flag = 1;//两对以上，直接结束
                    break;
                }
            }
        }
        if(flag == 1) {
            printf("No
");
            continue;
        }
        if(s[c1] == s[c2]&&t[c1] == t[c2]) {
            printf("Yes
");
        } else {//字母不同
            printf("No
");
        }
    }
    return 0;
}

 

 

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