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 F first
#define S second
#define pb push_back 
#define mp make_pair

#define ll long long 
#define ld long double 

#define vi vector <int>
#define vb vector <bool>
#define vc vector <char>
#define vll vector <ll>
#define vs vector <string>

#define pll pair < ll , ll >
#define pld pair < ld , ld >
#define pii pair < int , int >

#define vpld vector < pld >
#define vpii vector < pii >
#define vpll vector < pll >
#define vpci vector <pair <char,int> >

#define vvi vector < vector < int > >
#define vvll vector < vector < ll > >
#define vvld vector < vector < ld > >
#define vvb vector < vector < bool > >
#define vvpii vector < vector < pii > >
//vector<vector<int>> v(n, vector<int>(n));

#define all(v) (v).begin() , (v).end()
#define allrev(v) (v).rbegin() , (v).rend()

#define dist(x1 , y1 , x2 , y2) sqrt((x2 - x1)*(x2 - x1) + (y2 - y1)*(y2 - y1))

const int MOD = 1e9 + 7 ;

bool isprime (ll num) {
    for (ll i = 2 ; i * i <= num ; ++i) {
		if (num % i == 0) return false ;
	}
    return true;
}
 
ll gcd (ll a, ll b) {
    if (b == 0) return a ;
	else return gcd (b , a % b) ;
}
 
ll lcm (ll a, ll b) {
    return ((a / gcd(a , b)) * b) ;
}
 
ll cel (ll x1, ll y1) {
    if ((x1) <= 0) return (x1) / (y1) ;
    return (((x1) / (y1)) + (((x1) % (y1)) != 0)) ;
}

float squareRoot(float n) 
    { 
        float x = n; 
        float y = 1; 
        float e = 0.000001;
        while (x - y > e) { 
            x = (x + y) / 2; 
            y = n / x; 
        } 
        return x; 
    } 
    
void search(string pat, string txt)  
{  
    int M = pat.size();  
    int N = txt.size();  
    int i = 0;  
  
    while (i <= N - M)  
    {  
        int j;
        for (j = 0; j < M; j++)  
            if (txt[i + j] != pat[j])  
                break;  
        if (j == M) 
        {  
            cout << "Pattern found at index " << i << endl;  
            i = i + M;  
        }  
        else if (j == 0)  
            i = i + 1;  
        else
            i = i + j;  
    }  
}  
    
int power(int x, unsigned int y) 
{ 
    if (y == 0) 
        return 1; 
    else if (y%2 == 0) 
        return power(x, y/2)*power(x, y/2); 
    else
        return x*power(x, y/2)*power(x, y/2); 
} 
/*int INT(string s)
{
    return stoi(s);
}*/

int sum_of_digits(long long n) {
	int res = 0;
	while (n > 0) {
		res += n % 10;
		n /= 10;
	}
	return res;
}

void solve (int test_case) {
	int n,k;
	cin>>n>>k;
	string s1,s2;
	cin>>s1>>s2;
	vi v1(26),v2(26);
    for(int i=0;i<n;i++)
    {
        v1[s1[i]-'a']++;
        v2[s2[i]-'a']++;
    }
    bool flag=true;
    int p=0;
    for(int i=0;i<26;i++)
    {
        int c=p+v1[i]-v2[i];
        p=c;
        if(c<0||c%k) {flag=false; break;}
    }
	if(flag) cout<<"Yes"<<endl;
	else cout<<"No"<<endl;
}

int main() {
	ios_base :: sync_with_stdio (false) ;
	cin.tie(0) ;
	cout.tie(0) ;
	//freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
	int tc = 1 ;
	cin >> tc ;
	while (tc--) {
		solve (tc) ;
	}
	return 0 ;
}