Ashish has two strings a and b, each of length n, and an integer k. The strings only contain lowercase English letters.

He wants to convert string a into string b by performing some (possibly zero) operations on a.

In one move, he can either 

  * choose an index i (1 ≤ i≤ n-1) and swap a_i and a_{i+1}, or 
  * choose an index i (1 ≤ i ≤ n-k+1) and if a_i, a_{i+1}, …, a_{i+k-1} are all equal to some character c (c ≠ 'z'), replace each one with the next character (c+1), that is, 'a' is replaced by 'b', 'b' is replaced by 'c' and so on. 



Note that he can perform any number of operations, and the operations can only be performed on string a. 

Help Ashish determine if it is possible to convert string a into b after performing some (possibly zero) operations on it.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of each test case is as follows.

The first line of each test case contains two integers n (2 ≤ n ≤ 10^6) and k (1 ≤ k ≤ n).

The second line of each test case contains the string a of length n consisting of lowercase English letters.

The third line of each test case contains the string b of length n consisting of lowercase English letters.

It is guaranteed that the sum of values n among all test cases does not exceed 10^6.

Output

For each test case, print "Yes" if Ashish can convert a into b after some moves, else print "No".

You may print the letters of the answer in any case (upper or lower).

Example

Input


4
3 3
abc
bcd
4 2
abba
azza
2 1
zz
aa
6 2
aaabba
ddddcc


Output


No
Yes
No
Yes

Note

In the first test case it can be shown that it is impossible to convert a into b.

In the second test case,

"abba" \xrightarrow{inc} "acca" \xrightarrow{inc} … \xrightarrow{inc} "azza".

Here "swap" denotes an operation of the first type, and "inc" denotes an operation of the second type.

In the fourth test case,

"aaabba" \xrightarrow{swap} "aaabab" \xrightarrow{swap} "aaaabb" \xrightarrow{inc} … \xrightarrow{inc} "ddaabb" \xrightarrow{inc} … \xrightarrow{inc} "ddddbb" \xrightarrow{inc} … \xrightarrow{inc} "ddddcc".
#include <bits/stdc++.h>

using namespace std;
#define debug(a) for(int i=0;i<a.size();i++) cout<<a[i]<<" "; cout<<"\n";
typedef long long ll;

int main(){
	ios_base::sync_with_stdio(false); cin.tie(nullptr);
	
	//뭐 최소횟수를 구하라는게 아니라 가능/불가능 여부니까 dp는 아니고
	//연산은 
	//1. 인접한 문자간의 교환 가능
	//2.연속된 k개의 같은 문자를 알파벳 상 다음 문자로 교환 가능.
	//연산1에 의해 정렬하고 시작해도 됨
	//비가역성. 알파벳 상 증가만 되지 감소는 안됨
	//그럼 prefixsum만 중요하다?
	//정렬하고 prefixsum of string1 < prefixsum of string2 이면 가능할듯?
	
	//아 근데 k개씩 연속해야 하네..
	//브루트포스로 알파벳 오름차순으로 string1에서 줄이고 string2에서 맞춰나가자. 그래서 되면 되는거고 아니면 안되는거일듯 
	
	int Q; cin>>Q;
	while(Q--){
		int n,k; cin>>n>>k;
		string src,dst; cin>>src>>dst;
		
		vector<int> cnt(26,0),cnt2(26,0);
		for(int i=0;i<n;i++) cnt[src[i]-'a']++,cnt2[dst[i]-'a']++;
		
		//debug(cnt);
		//debug(cnt2);
		
		bool possible=true;
		for(int i=0;i<25;i++){		//'a'~'y'('z'전까지) 
			if(cnt[i]>cnt2[i]){
				int diff= cnt[i]-cnt2[i];
				if(diff%k){		
					possible=false; break;
				}else{
					cnt[i]-=diff;
					cnt[i+1]+=diff;
				}
			}else if(cnt[i]==cnt2[i]){
				continue;
			}else{		//impossible
				possible=false;
				break;
			}
		}
		
		if(cnt['z'-'a']!=cnt2['z'-'a']) possible=false;
		
		if(possible) cout<<"YES\n"; else cout<<"NO\n";
		
	} 
			
}