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 Adding_Powers(n, k, array):
    P=[]                                     #Potencias usadas
    for i in range(n):                      
         e = array[i]
         m = Max_Power(e, k)                # Encontrar el K^(m)

         while array[i] >0 and m>=0:
             if array[i] >= k**m:
                array[i] -= k**m              # Resto K^(m)      
                if P.count(m)==1: return "NO" #No puede haber potencias repetidas
                P.append(m) 
             m-=1

    if max(array) >0:
        return "No"             
    else:
        return "Yes"
#Encuentra la mayor potencia sumada
def Max_Power(M, k):

    m = 0
    while k**m < M:               # Encontrar K^(m+1)
        m += 1
    if(k**m > M):
        m -= 1                    # Encontrar el K^(m)
    return m

#Toma los casos
for l in range(int(input())):
    n, k = map(int, (input().split()))
    array= list(map(int, (input().split())))
    print(Adding_Powers(n, k, array))
