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;

typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
double pi = acos(-1);
#define tezi        ios_base::sync_with_stdio(false);\
                    cin.tie(0);\
                    cout.tie(0);
#define FOR(i,a,b)  for(int i=a;i<b;i++)
#define FORR(i,b,a) for(int i=b;i>=a;i--)
#define fill(a,b)   memset(a,b,sizeof(a))
#define _time_      1.0 * clock() / CLOCKS_PER_SEC
#define fi          first
#define se          second
#define mp          make_pair 
#define pb          push_back
#define all(a)      a.begin(),a.end()
#define rev(a)      reverse(all(a))
#define reva(a,n)   reverse(a,a+n)
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());

ll powerm(ll a,ll b,ll m){
    if(b==0) return 1;
    if(b&1) return a*powerm(a*a%m,b/2,m)%m;
    return powerm(a*a%m,b/2,m);
}
ll power(ll a,ll b){
    if(b==0) return 1;
    if(b&1) return a*power(a*a,b/2);
    return power(a*a,b/2);
}
ll sqroot(ll number){ 
    ll x = sqrt(number);
    while(x*x<number)x++;
    while(x*x>number)x--;
    return x;
}
ll cbroot(ll number){
    ll x = cbrt(number);
    while(x*x*x<number)x++;
    while(x*x*x>number)x--;
    return x;
}

const int maxn=1e5+5;
//bool prime[maxn];

int main(){
    tezi
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    //sieve
    //memset(prime,true,sizeof(prime)); 
    //for(ll p=2;p*p<maxn;p++)if(prime[p]==true)for(ll i=p*p;i<maxn;i+=p)prime[i]=false;
    int t;
    cin >> t;
    while(t--){
        int n,k;
        cin >> n >> k;
        string a,b;
        cin >> a >> b;
        vector<int> cnt1(27),cnt2(27);
        FOR(i,0,n){
            cnt1[b[i]-'a']++;
            cnt2[a[i]-'a']++;
        }
        int prev=0;
        bool ok=1;
        for(int i=0;i<26;i++){
            cnt2[i]+=prev;
            if(cnt2[i]<cnt1[i]){
                ok=0;
                break;
            }else{
                int ch = cnt2[i]-cnt1[i];
                if(ch%k){
                    ok=0;
                    break;
                }else{
                    prev=ch;
                }
            }
        }
        cout << (ok?"Yes\n":"No\n");
    }
    return 0;
}


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*/