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 JAVA):

Solution:


 //   * * * its fun to do the impossible * * *   //
 import java.io.BufferedReader;
 import java.io.IOException;
 import java.io.InputStreamReader;
 import java.util.*;
  
 
 public class StringEquality {
    
     public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        int t = sc.nextInt();
        while(t-->0){
            int n = sc.nextInt();
            int k = sc.nextInt();

            String a = sc.nextLine();
            String b = sc.nextLine();

            int c[] = new int[256];

            for(int i=0;i<n;i++){
                c[a.charAt(i)]++;
                c[b.charAt(i)]--;
            }

            // a -> z == 97 -> 122

            for(int i=97;i<123;i++){
                if(c[i] < 0  && c[i] % k == 0){
                    inner : for(int j=i-1;j>96;j--){
                        if(c[j] % k == 0){
                            int min = Math.min(-1*c[i] , c[j]);
                            c[i] += min;
                            c[j] -= min;
                            if(c[i] == 0){
                                break inner;
                            }
                        }
                    }
                }
            }

            boolean is = true;
            for(int i=97;i<123;i++){
                if(c[i] != 0){
                    is = false;
                    break;
                }
            }

            if(is)
                System.out.println("Yes");
            else
                System.out.println("No");
            
        }
         
 
     }
 
 
 
 
 
 
  
     // Use this instead of Arrays.sort() on an array of ints. Arrays.sort() is n^2
     // worst case since it uses a version of quicksort. Although this would never
     // actually show up in the real world, in codeforces, people can hack, so
     // this is needed.
     static void ruffleSort(int[] a) {
         //ruffle
         int n=a.length;
         Random r=new Random();
         for (int i=0; i<a.length; i++) {
             int oi=r.nextInt(n), temp=a[i];
             a[i]=a[oi];
             a[oi]=temp;
         }
         
         //then sort
         Arrays.sort(a);
     }
     
     // Use this to input code since it is faster than a Scanner
     static class FastScanner {
         BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
         StringTokenizer st=new StringTokenizer("");
         String next() {
             while (!st.hasMoreTokens())
                 try {
                     st=new StringTokenizer(br.readLine());
                 } catch (IOException e) {
                     e.printStackTrace();
                 }
             return st.nextToken();
         }
         
         int nextInt() {
             return Integer.parseInt(next());
         }
 
         double nextDouble() {
             return Double.parseDouble(next());
         }
  
         String str = "";
  
         String nextLine() {
             try {
                 str = br.readLine();
             } catch (IOException e) {
                 e.printStackTrace();
             }
             return str;
         }
  
         int[] readArray(int n) {
             int[] a = new int[n];
             for (int i = 0; i < n; i++)
                 a[i] = nextInt();
             return a;
         }
  
         long[] readLongArray(int n) {
             long[] a = new long[n];
             for (int i = 0; i < n; i++)
                 a[i] = nextLong();
             return a;
         }
         long nextLong() {
             return Long.parseLong(next());
         }
     }
     // use this to find the index of any element in the array +1 /// 
     // returns an array that corresponds to the index of the i+1th in the array a[]
     // runs only for array containing different values enclosed btw 0 to n-1
     static int[] indexOf(int[] a) {
         int[] toRet=new int[a.length];
         for (int i=0; i<a.length; i++) {
             toRet[a[i]]=i+1;
         }
         return toRet;
     }
 
     static int gcd(int a, int b) {
         if (b==0) return a;
         return gcd(b, a%b);
     }
 
     //generates all the prime numbers upto n
     static void sieveOfEratosthenes(int n , ArrayList<Integer> al)
     {
         boolean prime[] = new boolean[n + 1];
         for (int i = 0; i <= n; i++)
             prime[i] = true;
  
         for (int p = 2; p * p <= n; p++) 
         {
             if (prime[p] == true) 
             {
                 for (int i = p * p; i <= n; i += p)
                     prime[i] = false;
             }
         }
         for (int i = 2; i <= n; i++)
         {
             if (prime[i] == true)
                 al.get(i);
         }
     }
     static final int mod=100000000 + 7;
     
     //fastPow
     static long fastPow(long base, long exp) {
         if (exp==0) return 1;
         long half=fastPow(base, exp/2);
         if (exp%2==0) return mul(half, half);
         return mul(half, mul(half, base));
     }
 
     //multiply two long numbers
     static long mul(long a, long b) {
         return a*b%mod;
     }
 
     static int nCr(int n, int r)
     {
            return fact(n) / (fact(r) *
                 fact(n - r));
     }
     
     // Returns factorial of n
     static int fact(int n)
     {
         int res = 1;
         for (int i = 2; i <= n; i++)
             res = res * i;
         return res;
     }
      
     // to generate the lps array
     // lps means longest preffix that is also a suffix
     static void generateLPS(int lps[] , String p){
         int l = 0;
         int r = 1;
    
         while(l < p.length() && l < r && r < p.length()){
             if(p.charAt(l) == p.charAt(r)){
                 lps[r] = l + 1;
                 l++;
                 r++;
             }
             else{
                 if(l > 0)
                     l = lps[l - 1];
                 else
                     r++;    
             }
         }
     }
      
 
     //count sort --> it runs in O(n) time but compromises in space
     static ArrayList<Integer> countSort(int a[]){
         int max = Integer.MIN_VALUE;
         for(int i=0;i<a.length;i++){
             max = Math.max(max , a[i]);
         }
 
         int posfre[] = new int[max+1];
         boolean negPres = false;
         for(int i=0;i<a.length;i++){
             if(a[i]>=0){
                 posfre[a[i]]++;
             }
             else{
                 negPres = true;
             }
         }
         ArrayList<Integer> res = new ArrayList<>();
         if(negPres){
             int min = Integer.MAX_VALUE;
             for(int i=0;i<a.length;i++){
                 min = Math.min(min , a[i]);
             }
 
             int negfre[] = new int[-1*min+1];
             for(int i=0;i<a.length;i++){
                 if(a[i]<0){
                     negfre[-1*a[i]]++;
                 }
             }
 
             for(int i=min;i<0;i++){
                 for(int j=0;j<negfre[-1*i];j++){
                     res.add(i);
                 }
             }
             for(int i=0;i<=max;i++){
                 for(int j=0;j<posfre[i];j++){
                     res.add(i);
                 }
             }
             return res;
         }
         for(int i=0;i<=max;i++){
             for(int j=0;j<posfre[i];j++){
                 res.add(i);
             }
         }
         return res;
     }
 
     // write 
 
 }