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 power(n,k):
    aa=0
    while(n%k==0):
        n/=k
        aa+=1
    return aa
 
def xxx(n,k):
    ans=[]  
    x=1
    land=n
    if(n==1):
        return [0]
    ans=[]
    x=1
    if(n%k!=0):
        n-=1
        ans.append(0)
    while x:
        temp=power(n,k)
        ans.append(temp)
        zz=n
        n/=(k**temp)
        if(zz==n):
            break
        if(n==1):break
        if(n<k):break
        n-=1
    axy=[x for x in ans]
    for i in range(0,len(axy)-1):
        for j in range(i+1,len(ans)):
            axy[j]+=ans[i]
    
    test=0
    for j in axy:
        test+=k**j
    if(test!=land):return -1
    if(len(set(axy))!=len(axy)):return -1
    return axy

for _ in range(int(input())):
    n,k=map(int,input().split())
    arr=list(map(int,input().split()))
    y=set(arr)
    y=list(y)
    if(y[0]==0 and len(y)==1):
        print('YES')
        continue
    else:
        ans_l=[]
        flag=0
        for i in range(0,n):
            if( arr[i]==0):continue
            if(flag):break
            z=xxx(arr[i],k)
            if(z==-1):
                flag=1
                break
            for m in z:
                if(m in ans_l):
                    flag=1
                    break
            ans_l+=z
        if(flag):print('NO')
        else:print('YES')