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:
/*Every winner is a loser he just tried one last time*/

#include<bits/stdc++.h>
using namespace std;
const long long mod = 1e9+7;
const long long mx  = 1e5+5;

#define inf 			9e18
#define zero 			-9e18
#define PI   			acos(-1.0)  // 3.1415926535897932
#define Case(J) 		printf("Case %lld: %lld\n",++count,J); ///print case
#define preci           cout<<fixed<<showpoint<<setprecision(20);
#define max3(a,b,c) 		max(a,max(b,c))
#define min3(a,b,c) 		min(a,min(b,c))
#define GCD(a,b) 		__gcd(a,b)
#define LCM(a,b) 		(a*(b/__gcd(a,b) ))
#define MP 			make_pair
#define pb 			push_back
#define ppb 			pop_back()
#define pf 			push_front
#define ppf 			pop_front()
#define rev(v) 			reverse(v.begin(),v.end());
#define srt(v) 			sort(v.begin(),v.end());
#define grtsrt(v) 		sort(v.begin(),v.end(),greater<ll>());
#define hellyeah 		exit(0);
#define file 			freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);
#define vmax(V) 		*max_element(V.begin(),V.end());
#define vmin(V) 		*min_element(V.begin(),V.end());
#define	debug(a) 		cout<<"*"<<a<<"$"<<endl;
#define	debug2(a,b) 		cout<<"$"<<a<<" "<<b<<"$"<<endl;
#define	debug3(a,b,c) 		cout<<"$"<<a<<" "<<b<<" "<<c<<"$"<<endl;
#define SET(X,D_type)		memset(X, D_type, sizeof(X))
#define groot(A)		{cout<<A<<endl;return;}
#define vins(V)			srt(V)V.resize(unique(V.begin(),V.end())-V.begin());
#define check           	cout<<"Avengers Assemble"<<endl;
#define	lol 			cout<<endl;
#define meem(X) 		cout<<(X?"Yes":"No")<<endl;
#define joker(V) 		for(auto X:V)cout<<X<<" ";cout<<endl;
#define papiya(Mp) 		for(auto X:Mp)cout<<X.first<<" "<<X.second<<endl;
#define lp(i,a) 		for(ll i=0; i<a;i++)
#define lp1(i,a,b) 		for(ll i=a; i<=b;i++)
#define rlp(i,b,a) 		for(ll i=(b);i>=(a);--i) 
#define Mat(mat,N,M)		lp(i,N){lp(j,M)cout<<mat[i][j]<<" \n"[j==M-1];}
typedef long long ll;
typedef long double ld;
void Loser(){ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);}
ll fact(ll N){return (N<=1ll?1ll:N*fact(N-1ll));}
long long ceil(ll A,ll B){if(A%B==0)return A/B;else return (A/B)+1ll;}
long long middle(ll a, ll b, ll c) { if ((a <= b && b <= c) || (c <= b && b <= a)) return b;else if ((b <= a && a <= c) || (c <= a && a <= b))return a;else return c; } 
bool ispalindrom(string s) { ll i=0, j=s.size()-1;while(i<j){if(s[i]!=s[j])return 0;i++,j--;}return 1;}
//sort(v.begin(),v.end(),as_second);
bool as_second(const pair<ll,ll> &a, const pair<ll,ll> &b) {return (a.second < b.second); }//sort the vector pair in assending order according to second element
bool ds_first(const pair<ll,ll> &a, const pair<ll,ll> &b){ return (a.first > b.first);}//sort the vector pair in decending order according to first element
bool ds_second(const pair<ll,ll> &a, const pair<ll,ll> &b) {return a.second>b.second;}//sort the vector pair in decending order according to second element

typedef deque<ll> Dq;
typedef set<ll> St;
typedef map<ll,ll> M;
typedef vector<ll> V;
typedef vector<V> VV;
typedef vector<pair<ll,ll>> VP;
typedef priority_queue<ll,vector<ll>,greater<ll>> Pq;
string str,str1,str2,str3,str4,str5,str6,str7;
char ch,ch1,ch2,ch3;
bool flag,flag1,flag2;

long long a,b,c,d,e,f,g,h,l,i,j,k,m,n,o,p,q,r,s,t,u,v,w,x,y,z,test,cnt,cnt1,cnt2,cnt3,cnt4,sum,sum1,sum2,sum3,ans,ans1,ans2,ans3,ma,ma1,ma2,ma3,mi,mi1,mi2,mi3,temp,temp1,temp2,temp3,temp4;
//long double a,b,c,d,e,f,g,h,l,i,j,k,m,n,o,p,q,r,s,t,u,v,w,x,y,z,cnt,cnt1,cnt2,cnt3,cnt4,sum,sum1,sum2,sum3,ans,ans1,ans2,ans3,ma,ma1,ma2,ma3,mi,mi1,mi2,mi3,temp,temp1,temp2,temp3,temp4;

/*-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/

void SectumSempra()
{
	int n, k;
	cin >> n >> k;
	string s, t;
	cin >> s >> t;
	int freq1[26] = {};
	int freq2[26] = {};
	for (char c : s)
		++freq1[c - 'a'];
	for (char c : t)
		++freq2[c - 'a'];
		
	bool f = true;
	for (int i = 0; i < 26; ++i) {
		if (freq1[i] < freq2[i] or (freq1[i] - freq2[i]) % k)
			f = false;
		else
			freq1[i + 1] += freq1[i] - freq2[i];
	}
	cout << (f ? "Yes" : "No") << endl;
	lol
}

int main()
{
	Loser();
	//test=1;
	cin>>test;
	while(test--)
	{
		SectumSempra();
	}
	return 0;
}