Codeforces Round #598 (Div. 3) F. Equalizing Two Strings 构造

F. Equalizing Two Strings

You are given two strings s and t both of length n and both consisting of lowercase Latin letters.

In one move, you can choose any length len from 1 to n and perform the following operation:

Choose any contiguous substring of the string s of length len and reverse it;

at the same time choose any contiguous substring of the string t of length len and reverse it as well.

Note that during one move you reverse exactly one substring of the string s and exactly one substring of the string t.

Also note that borders of substrings you reverse in s and in t can be different, the only restriction is that you reverse the substrings of equal length. For example, if len=3 and n=5, you can reverse s[1…3] and t[3…5], s[2…4] and t[2…4], but not s[1…3] and t[1…2].

Your task is to say if it is possible to make strings s and t equal after some (possibly, empty) sequence of moves.

You have to answer q independent test cases.

Input

The first line of the input contains one integer q (1≤q≤104) — the number of test cases. Then q test cases follow.

The first line of the test case contains one integer n (1≤n≤2⋅105) — the length of s and t.

The second line of the test case contains one string s consisting of n lowercase Latin letters.

The third line of the test case contains one string t consisting of n lowercase Latin letters.

It is guaranteed that the sum of n over all test cases does not exceed 2⋅105 (∑n≤2⋅105).

Output

For each test case, print the answer on it — "YES" (without quotes) if it is possible to make strings s and t equal after some (possibly, empty) sequence of moves and "NO" otherwise.

Example

input

4

4

abcd

abdc

5

ababa

baaba

4

asdf

asdg

4

abcd

badc

output

NO

YES

NO

YES

题意

现在给你两个字符串，你可以进行若干次操作。

每次操作需要在每个字符串都选择出长度为len的一个区间，然后将这个区间的字符都进行翻转。

问你进行若干次操作后，这俩字符串能变成一样的吗？

题解

按照这个顺序进行判断：

1. 如果两个字符串存在不同的字符，那么肯定是NO
2. 如果某个字符串存在两个相同的字符，那么一定是YES，因为可以就在这两个字符中进行无限次的翻转
3. 如果两个字符串的逆的奇偶性相同，那么一定是YES

第三个怎么理解呢？在判断1和2之后，我们得到的一定是一个排列，问题就变成你可以翻转若干次，两个排列能否相同。

我们考虑我们同时翻转相同长度的，我们排列的逆一定会发生奇偶性的变化，那么如果一开始奇偶性就不同，那么不管怎么翻转，都不会相同。

代码

#include<bits/stdc++.h>
using namespace std;

int n;
string s1,s2,S1,S2;
int Count(string s){
	int num=0;
	for(int i=0;i<s.size();i++){
		for(int j=0;j<i;j++){
			if(s[j]>s[i])
				num++;
		}
	}
	return num;
}
void solve(){
	cin>>n>>S1>>S2;
	s1=S1;s2=S2;
	sort(s1.begin(),s1.end());
	sort(s2.begin(),s2.end());
	for(int i=0;i<n;i++){
		if(s1[i]!=s2[i]){
			puts("NO");
			return;
		}
	}
	for(int i=1;i<n;i++){
		if(s1[i]==s1[i-1]){
			puts("YES");
			return;
		}
		if(s2[i]==s2[i-1]){
			puts("YES");
			return;
		}
	}

	if(Count(S1)%2==Count(S2)%2){
		puts("YES");
	}else{
		puts("NO");
	}
	return;
}
int main(){
	int t;
	scanf("%d",&t);
	while(t--)solve();
}