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:
// start fold
#include <bits/stdc++.h>
#ifdef LOCAL
#include <sys/resource.h>
#include <sys/time.h>
#define MAX_FILE_SIZE 536870912
#define CPU_TIME 2
#endif
#define endl '\n'
using namespace std;
using ull = unsigned long long;
using ll = long long;
using ld = long double;

#define in cin
#define out cout

typedef int                     integer;
#define int                     long long
typedef pair<int, int>          pii;
#define vii                     vector<pair<int,int>>
#define vi                      vector<int>
#define vvi                     vector<vector<int>>
#define rep(i, a, b)            for(int i = a; i < (b); i++)
#define s(x)                    in>>x
#define s2(x,y)                 in>>x>>y
#define s3(x,y,z)               in>>x>>y>>z
#define s4(x,y,z,w)             in>>x>>y>>z>>w
#define p(x)                    out<<(x)<<endl
#define p2(x,y)                 out<<(x)<<' '<<(y)<<endl
#define p3(x,y,z)               out<<(x)<<' '<<(y)<<' '<<(z)<<endl
#define p4(x,y,z,w)             out<<(x)<<' '<<(y)<<' '<<(z)<<' '<<(w)<<endl
#define p1d(a,n)                for(int ix=0;ix<n-1;ix++) out<<a[ix]<<' '; out<<a[n-1]<<'\n';
#define p2d(a,n,m)              for(int ix=0;ix<n;ix++){ for(int jx=0;jx<m;jx++) out<<a[ix][jx]<<" "; out<<endl;}
#define pl(x)                   out<<(x)
#define pll(x)                  out<<(x)<<' '
#define range(v)                v.begin(),v.end()
#define all(x)                  begin(x), end(x)
#define sz(x)                   (int)(x).size()
#define rst(x)                  memset(x, -1, sizeof(x))

#define EACH(x, a) for (auto& x: a)
template<class T> void read_v(T& x, std::istream& in) {
    in >> (x);
}
template<class A> void read_vec(vector<A>& x, std::istream& in) {
    EACH(a, x)
        read_v(a, in);
}

void __print(integer x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (const auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}

#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "] = ["; _print(__VA_ARGS__)
#else
#define debug(...) 42
#endif

int iceildiv(const int a, const int b){ if(a==0){return 0;} return (1 + ((a - 1)/b));}
bool sortpairbysec(const pair<int,int> &a,const pair<int,int> &b){return (a.second < b.second);}
template <typename T> T factmod(T a, T modulus){ int res = 1, i; for (i = 2; i <= a; i++){ res *= i; res%=modulus;} return res; }
template <typename T> int powmod(T a, T b, const int m){ int ans=1; while(b>0){if(b&1){ans*=a;ans%=m;}a*=a;a%=m;b>>=1;} return ans; }
template <typename T> int power(T a, T b){ int ans=1; while(b>0){if(b&1){ans*=a;}a*=a;b>>=1;}return ans;}
template <typename T> void setmin(T& a, T b) { if (b < a) a = b; }
template <typename T> void setmax(T& a, T b) { if (b > a) a = b; }
// end fold

void solve(){
    int n, k; s2(n, k);
    string st1, st2; s2(st1, st2);
    debug(st1);
    debug(st2);
    vi cnt1(26, 0), cnt2(26, 0);
    for(char c: st1) cnt1[c-'a']++;
    for(char c: st2) cnt2[c-'a']++;
    for(int i = 0; i < 25; i++){
        int diff = cnt2[i] - cnt1[i];
        if(diff == 0) continue;
        if(diff > 0){ p("No"); return; }

        if(diff < 0){
            if(abs(diff) >= k){
                cnt1[i + 1] += k * (abs(diff)/k);
            }
        }
    }
    if(cnt1[25] == cnt2[25]){
        p("Yes");
    }else{
        p("No");
    }
    return;
}

// start fold
integer main() {
#ifdef LOCAL
        rlimit rlim, rlim_time;
        if(getrlimit(RLIMIT_FSIZE, &rlim) || getrlimit(RLIMIT_CPU , &rlim_time))  
            return 1 ; 
        rlim.rlim_cur = MAX_FILE_SIZE; // maximum file size (in bytes) that your program will be able to create
        rlim_time.rlim_cur = CPU_TIME; // maximum allowed runtime for your program
        if(setrlimit(RLIMIT_FSIZE, &rlim) || setrlimit(RLIMIT_CPU , &rlim_time))
            return 2;
#endif
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int test_cases = 1;
    cin >> test_cases;
    while(test_cases--){
        solve();
    }
    return 0;
}
// end fold