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 conv(num,b):
    if num<b:
        return str(num)
    else:
        return conv(num//b,b) +" "+ str(num%b)
t=int(input())
for _ in range(t):
    n,k=tuple(map(int,input().split(" ")))
    num=list(map(int,input().split(" ")))
    sum=0
    flag=0
    prev_list=[]
    for i in num:
        if(i==0):
            pass
        else:
            convert=list(map(int,conv(i,k).split(" ")))
            convert.reverse()
            
            for j in range(len(convert)):
                if(convert[j]>1):
                    flag=1
                    break
                if(j<len(prev_list)):
                    if(prev_list[j]==1 and convert[j]==1):
                        flag=1
                        break
                    elif(prev_list[j]==0 and convert[j]==1):
                        prev_list[j]=1
                else:
                    prev_list.append(convert[j])
        if(flag==1):
            break
    if(flag==0):
        print("YES")
    else:
        print("NO")