Problem: 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".

Solution (in CPP):

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

#define ll long long
#define ld long double

#define uset unordered_set
#define umap unordered_map

#define ar array
#define vt vector
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()

#define FOR_B(i, a, b, x) for(int i=(a); (x)>0?i<(b):i>(b); i+=(x))
#define FOR_1(b) FOR_B(i, 0, b, 1)
#define FOR_2(i, b) FOR_B(i, 0, b, 1)
#define FOR_3(i, a, b) FOR_B(i, a, b, 1)
#define FOR_4(i, a, b, x) FOR_B(i, a, b, x)
#define FOR_Q(a, b, c, d, e, ...) e
#define FOR_G(...) FOR_Q(__VA_ARGS__, FOR_4, FOR_3, FOR_2, FOR_1)
#define FOR(...) FOR_G(__VA_ARGS__)(__VA_ARGS__)
#define EACH(x, a) for(auto& x : a)

template<class T> string to_string(T v) {
	bool f=0;
	string res;
	EACH(x, v) {
		res+=(f?" ":"");
		f=1;
		res+=to_string(x);
	}
	return res;
}
string to_string(const char c) {
	return string(1, c);
}
string to_string(const bool b) {
	return b?"true":"false";
}
string to_string(const char* s) {
	return string(s);
}
string to_string(const string s) {
	return s;
}
string to_string(const vt<bool> v) {
	string res;
	FOR(sz(v))
		res+=(char)('0'+v[i]);
	return res;
}

template<class T> void read(T& x) {
	cin >> x;
}
template<class T> void read(vt<T>& a) {
	EACH(x, a)
		read(x);
}
template<class T, size_t S> void read(ar<T, S>& a) {
	EACH(x, a)
		read(x);
}
template<class T, class... O> void read(T& x, O&... o) {
	read(x);
	read(o...);
}

template<class T> void write(const T& x) {
	cout << to_string(x);
}
template<class T, class... O> void write(const T& x, const O&... o) {
	write(x);
	//if(sizeof...(o))
		//write(" ");
	write(o...);
}
void write() {
	write("\n");
}

template<class T, class U> void vti(vt<T>& v, U x, size_t n) {
	v=vt<T>(n, x);
}
template<class T, class U> void vti(vt<T>& v, U x, size_t n, size_t m...) {
	v=vt<T>(n);
	EACH(i, v)
		vti(i, x, m);
}



int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);

	int t;
	read(t);
	while(t--) {
		int n, k;
		string a, b;
		read(n, k, a, b);
		vt<int> da(26), db(26);
		FOR(n)
			++da[a[i]-'a'],
			++db[b[i]-'a'];
		bool ok=1;
		//write("da: ", da, "\n");
		//write("db: ", db, "\n");
		for(int i=0; i<26; ++i) {
			int l=da[i]/k;
			for(int j=i; j<26; ++j) {
				int l2=db[j]/k;
				int m=min(l, l2);
				da[i]-=m*k;
				db[j]-=m*k;
				l-=m;
			}
		}
		FOR(26) if(da[i]^db[i]) ok=0;
		//write("da: ", da, "\n");
		//write("db: ", db, "\n");
		write(ok?"YES":"NO");
		write();
	}
}