Suppose you are performing the following algorithm. There is an array v_1, v_2, ..., v_n filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can: 

  * either choose position pos (1 ≤ pos ≤ n) and increase v_{pos} by k^i; 
  * or not choose any position and skip this step. 



You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (v_j = a_j for each j) after some step?

Input

The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.

The first line of each test case contains two integers n and k (1 ≤ n ≤ 30, 2 ≤ k ≤ 100) — the size of arrays v and a and value k used in the algorithm.

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{16}) — the array you'd like to achieve.

Output

For each test case print YES (case insensitive) if you can achieve the array a after some step or NO (case insensitive) otherwise.

Example

Input


5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810


Output


YES
YES
NO
NO
YES

Note

In the first test case, you can stop the algorithm before the 0-th step, or don't choose any position several times and stop the algorithm.

In the second test case, you can add k^0 to v_1 and stop the algorithm.

In the third test case, you can't make two 1 in the array v.

In the fifth test case, you can skip 9^0 and 9^1, then add 9^2 and 9^3 to v_3, skip 9^4 and finally, add 9^5 to v_2.
import sys
import math
import bisect
from sys import stdin, stdout
from math import gcd, floor, sqrt, log2, ceil
from collections import defaultdict as dd
from bisect import bisect_left as bl, bisect_right as br
from bisect import insort
from collections import Counter
from collections import deque
from heapq import heappush,heappop,heapify
from itertools import permutations,combinations
mod = int(1e9)+7
mod = 998244353
 
 
ip = lambda : int(stdin.readline())
inp = lambda: map(int,stdin.readline().split())
ips = lambda: stdin.readline().rstrip()
out = lambda x : stdout.write(str(x)+"\n")

t = ip()
for _ in range(t):
    n,k = inp()
    arr = list(inp())
    bit = [False]*55
    ch = sum(arr)
    flag = True
    for i in arr:
        ch = i
        temp = []
        while ch>0:
            rem = ch%k
            ch = ch//k
            if rem == 0 or rem == 1:
                temp.append(rem)
            else:
                flag = False
                break
        if flag:
            for i in range(len(temp)):
                if temp[i] == 1:
                    if bit[i]:
                        flag = False
                        break
                    else:
                        bit[i] = True
        else:
            break
            
    if flag:
        ans = "YES"
    else:
        ans = "NO"
    out(ans)
        
        
            
    
        
            
        
    
            
        
        
    

    
    
        
            
        
            
        
        
            
            
        
    
        
                    
                 
    
    
    
    
    
    
    
            
                
                
    
        
    
            
    
                    
                    
            
    
            
                
                
            
    
        
        
                    
        
        
            
            
            
        

        
    
        
    
            
    
    
        
                
        
        
            
                        
                
                    
                
                    
                
        
    
        
    
    
    
                        
            
            
    
            
            
        
            
                    
            
                    
                    
                    
                    
                    
                    
                    
    
        
    
    
    
    
    
            
        
                
        
            
        
    
    
    
            
            
        
    
            
            
        
                
                
        
    
        
        
    
    
            

        
        
        

            
            
        
            
        
        
        
            
        
    
            
        
                
        
        
            
            
    
                
        
                
            
            
        
    
        
    
                
        
    
    
    
    
    
    
            

    
            
            
            
        
    

            
        
