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 answer(k, target) :
    ithSteps = []
    maxStep = 1

    for i in range(len(target)) :
        ithStep = []
        if target[i] < k :
            if target[i] != 0 and target[i] != 1 :
                return "NO"
            
            ithStep.append(target[i])
            ithSteps.append(ithStep)
        else :
            if target[i]%k != 0 and (target[i]-1)%k != 0 :
                return "NO"
            
            while(True) :
                rest = target[i]%k

                if rest != 0 and rest != 1 :
                    return "NO"
                target[i] = int((target[i] - rest)/k)
                ithStep.append(rest)

                if target[i] == 1 :
                    ithStep.append(1)
                    ithSteps.append(ithStep)
                    if maxStep < len(ithStep) :
                        maxStep = len(ithStep)
                    break

    for i in range(maxStep) :
        colSum = 0
        for j in range(len(ithSteps)) :
            if len(ithSteps[j]) > i :
                colSum += ithSteps[j][i]
                if colSum > 1 :
                    return "NO"
    
    return "YES"

testCount = int(input())
for _ in range(testCount) :
    n, k = map(int, input().split())
    target = list(map(int, input().split()))
    print(answer(k, target))