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.
# 1
# 2 2
# 1099511627776 1099511627775

t = int(input())
from math import log, ceil,floor

# print(9**5)
# print(log(1099511627775, 2))

def solve(no, k, ans):
    # print("no = ", no)
    global answer
    try:
        z = log(no, k)
        # print(z)
        # print(k**ceil(z))
        if k**(ceil(z)) > no:
            if k**(ceil(z) - 1) > no:
                z = ceil(z) - 2
            else:
                z = ceil(z) - 1
        else:
            z = ceil(z)
        # print(k**z)
        # if k**z < no:
            # z = (int(z))
        # if ceil(z) - z < 0.0000000000001:
            # z = ceil(z)
        z = (int(z))
        if z in ans:
            answer = False
            return
        else:
            ans[int(z)] = 1            
            solve(no - k**(int(z)), k, ans)
    except:
        pass

# t = 1    

for i in range(t):
    n, k = list(map(int,input().split()))
    # n = 1
    # k = 2
    # arr = list(map(int,"9007199254740991".split()))
    # arr = [1099511627776, 1099511627775]
    # arr = [1125899906842623, 562949953421312]
    arr = list(map(int,input().split()))
    ans = {}
    answer = True
    for i in arr:
        solve(i, k, ans)
    if answer:
        print("YES")
    else:
        print("NO")