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 PYTHON3):

Solution:
from collections import Counter
import string
import math
import sys
# sys.setrecursionlimit(10**6) 
from fractions import Fraction
from itertools import product
def array_int():
    return [int(i) for i in sys.stdin.readline().split()]
def vary(arrber_of_variables):
    if arrber_of_variables==1:
        return int(sys.stdin.readline())
    if arrber_of_variables>=2:
        return map(int,sys.stdin.readline().split()) 
def makedict(var):
    return dict(Counter(var))
# i am noob wanted to be better and trying hard for that
def printDivisors(n): 
    divisors=[]  
    # Note that this loop runs till square root 
    i = 1
    while i <= math.sqrt(n): 
          
        if (n % i == 0) : 
              
            # If divisors are equal, print only one 
            if (n//i == i) : 
                divisors.append(i) 
            else : 
                # Otherwise print both 
                divisors.extend((i,n//i)) 
        i = i + 1
    return divisors
def countTotalBits(num):
     binary = bin(num)[2:]
     return(len(binary))
def isPrime(n):
    # Corner cases 
    if (n <= 1) : 
        return False
    if (n <= 3) : 
        return True
  
    # This is checked so that we can skip  
    # middle five numbers in below loop 
    if (n % 2 == 0 or n % 3 == 0) : 
        return False
  
    i = 5
    while(i * i <= n) : 
        if (n % i == 0 or n % (i + 2) == 0) : 
            return False
        i = i + 6
  
    return True
mod=10**9+7
# def ncr(n,r):
#     if n<r:
#         return 0
#     if n==r:
#         return 1
#     numer=fact[n]
#     # print(numer)
#     denm=(fact[n-r]*fact[r])
#     # print(denm)
#     return numer*pow(denm,mod-2,mod)
# def dfs(node):
#     global graph,m,cats,count,visited,val
#     # print(val)
#     visited[node]=1
#     if cats[node]==1:
#         val+=1
    
    
#     # print(val)
#     for i in graph[node]:
#         if visited[i]==0:
#             z=dfs(i)
#             # print(z,i)
#             count+=z
#         val-=1
#         return 0
#     else:
#         return 1
    
    
# fact=[1]*(1001)
# c=1
# mod=10**9+7
# for i in range(1,1001):
# print(fact)
def comp(x):
#     fact[i]=(fact[i-1]*i)%mod
    return x[1]
def SieveOfEratosthenes(n): 
    # Create a boolean array "prime[0..n]" and initialize 
    #  all entries it as true. A value in prime[i] will 
    # finally be false if i is Not a prime, else true. 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
          
        # If prime[p] is not changed, then it is a prime 
        if (prime[p] == True): 
              
            # Update all multiples of p 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
      
    # Print all prime numbers 
    for p in range(2, n+1): 
        if prime[p]: 
            primes.append(p*p)
    primes=[]
# primes=[]
# SieveOfEratosthenes(2*(10**6))
def binary_search(arr, x):
    low = 0
    
    high = len(arr) - 1
    mid = 0
    while low <= high:
        mid = (high + low) // 2
  
        # Check if x is present at mid 
        if arr[mid] < x: 
            low = mid + 1
  
        # If x is greater, ignore left half 
        elif arr[mid] > x: 
            high = mid - 1
  
        # If x is smaller, ignore right half 
#     if val>m:
        else: 
            return mid 
  
    # If we reach here, then the element was not present 
    return -1
def lcm(a,b):
    return (a*b)//math.gcd(a,b)
mod=10**9+7
testCases=1
testCases=vary(1)
for _ in range(testCases):
    n,k=vary(2)
    a=(sys.stdin.readline())
    b=(sys.stdin.readline())
    rem=0
    for i in range(26):
        az=a.count(chr(97+i))
        az+=rem
        bz=b.count(chr(97+i))
        if az>=bz:
            rem=az-bz
            if (az-bz)%k==0:
                continue
            else:
                print('NO')
                break
        else:
            print('NO')
            break
    else:
        print('YES')