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.
def cs(n,k):
    a=n
    b=a
    C=[]
    d=0
    while(a!=0):
        if ((a%k)==0):
            a=a//k
            d=d+1

        else:
            C.append(d)
            a=b-(k**d)
            d=0
            b=a
    return C


T=int(input())
for _ in range(T):
    n,k=map(int,input().split())
    A=list(map(int,input().split()))
    A.sort()
    a=A[n-1]
    flag=0
    for i in range(n-1):
        if (A[i]!=0 and A[i]==A[i-1]):
            flag=1
            break

    if (A[n-1]==0):
        flag=0
        
    if (flag==0 and A[n-1]!=0):
            b=1
            ty=1
            while(b<A[n-1]):
                b=b*k
                ty=ty+1
            D=[0 for i in range(ty+1)]
            for i in range(n):
                E=cs(A[i],k)
                
                for j in E:
                    if (D[j]==1):
                        flag=1
                        break
                    else:
                        D[j]=1

                if (flag==1):
                    break
    if (flag==1):
        print("NO")

    else:
        print("YES")
            

    
        
                   
          
        
           
            
        
        

        
        
        
