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:
#pragma GCC optimize ("Ofast", "unroll-loops", "loop-nest-optimize")

#include <bits/stdc++.h>

#define REP(i, n) REPI(i, 0, n)
#define REP1(i, n) REPI(i, 1, n)
#define REPI(i, init, n) for(int i = init; i < n; i++)
#define MS(arr, val) memset(arr, val, sizeof(arr))
#define SZ(x) (x).size()
#define ALL(x) (x).begin(), (x).end()
#define EB emplace_back
#define UB upper_bound
#define LB lower_bound
#define FS first
#define SC second

#ifdef __WIN32__
inline size_t fread_unlocked(void *_buf, int _sz, int _cnt, FILE *_stream)
{
	return fread(_buf, _sz, _cnt, _stream);
}//fread_unlocked
#endif

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef long double LDB;
//typedef pair<int, int> P;

char buf[100000];

inline char gc()
{
	static char *st = buf, *ed = buf;

	if (st >= ed)
	{
		st = buf;
		ed = st + fread_unlocked(buf, 1, sizeof(buf), stdin);

		if (st == ed) return EOF;
	}//if

	return *(st++);
}//gc

template<class T>
inline bool rit(T &_a)
{
	char _c = gc();
	bool neg = false;
	for (; _c < '0' || _c > '9'; _c = gc())
	{
		if (_c == EOF) return false;
		if (_c == '-') neg = true;
	}//for

	for (_a = 0; _c >= '0' && _c <= '9'; _c = gc())
		_a = _a * 10 + _c - '0';

	if (neg) _a = -_a;

	return true;
}//rit

template<class S, class ...T>
inline bool rit(S &_a, T&... _b)
{
	if (!rit(_a)) return false;
	return rit(_b...);
}//rit

//============================================================================//

const int MAX_N = 0 + 10;
//const int INF8 = 0x3f;
//const int INF32 = 0x3f3f3f3f;
//const int MOD = 1e9 + 7;
//const int MOD = 46337; //maximum prime int for multiplying
//const LL MOD = 2999999929; //maximum prime LL for multiplying
//const LL INF64 = 0x3f3f3f3f3f3f3f3f;
//const DB PI = acos(-1);

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	
	int t;
	
	cin >> t;
	while (t--)
	{
		int n, k;
		string a, b;
		
		cin >> n >> k >> a >> b;
		
		int cntA[26] = {}, cntB[26] = {};
		
		for (char c : a) cntA[c - 'a']++;
		for (char c : b) cntB[c - 'a']++;
		
		bool flag = true;
		for (int i = 25; i >= 0 && flag; i--)
		{
			if (cntA[i] == cntB[i]) continue;
			else if (cntA[i] > cntB[i] || (cntB[i] - cntA[i]) % k) flag = false;
			else
			{
				for (int j = i - 1; j >= 0; j--)
				{
					int tmp = (cntB[i] - cntA[i]) / k;
				
					if (cntA[j] >= k)
					{
						cntA[i] = min(cntA[i] + cntA[j] / k * k, cntB[i]);
						cntA[j] = max(cntA[j] % k, cntA[j] - k * tmp);
					}//if
				}//for j
				
				if (cntA[i] < cntB[i]) flag = false;
			}//else
		}//for i
		
		if (flag) cout << "YES\n";
		else cout << "NO\n";
	}//while
	
	return 0;
}//main