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.
from math import log
from collections import defaultdict as dd
t=int(input())
while t:
    lol=0
    n,k=map(int,input().split())
    l=list(map(int,input().split()))    
    #l.sort()
    d=dd(int)
    for i in range(n):
        a=l[i]
        if(a%k!=0):
            if(d[0]==0):
                a-=1
                d[0]=1
            if(a%k!=0):
                lol=1
                break
        while a:
            if(a%k==0):
                cou=0
                aa=a
                while aa%k==0:
                    cou+=1
                    aa=aa//k
                if(d[cou]!=0):
                    lol=1
                    break
                a=a-k**cou
                d[cou]=1
            else:
                a-=1
                if(a%k!=0):
                    lol=1
                    break
        if(lol):
            break
            
    '''        
    lol=0
    d=dd(int)
    for i in range(n):
        a=l[i]
        if(a%k!=0 and d[0]==0):
            a-=1
            d[0]=1
            if(a%k!=0):
                lol=1
                break
        elif(a%k!=0 and d[0]==1):
            lol=1
            break
        #print(a)
        while a and lol==0:
            if(a%k==0):
                cou=0
                while a%k==0:
                    a=a//k
                    cou+=1
                if(d[cou]==1):
                    lol=1
                    
                d[cou]=1
            else:
                a=a-1
                if(a%k!=0):
                    lol=1
                if(a==0):
                    break
                    
        if(lol):
            break
    '''
    if(lol):
        print("NO")
    else:
        print("YES")
    t-=1
              