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 binary_pow(k, x):
    if (x == 1):
        return k
    if (x == 0):
        return 1
    else:
        if (x % 2 == 0):
            return binary_pow(k, x // 2) * binary_pow(k, x // 2)
        else:
            return k * (binary_pow(k, (x - 1) // 2)) * binary_pow(k, (x - 1) // 2)


def func(x):
    global k
    left = 0
    right = 100
    while (right - left > 1):
        mid = (left + right) // 2
        if binary_pow(k, mid) < x:
            left = mid
        else:
            right = mid
    ans = []
    for i in range(right, -1, - 1):
        v = x // binary_pow(k, i)
        ans.append(v)
        x -= (v * binary_pow(k, i))
    return ans[::-1]


t = int(input())
for i in range(t):
    n, k = map(int, input().split())
    a = [int(i) for i in input().split()]
    good = [0] * 200
    for i in range(n):
        x = func(a[i])
        for j in range(len(x)):
            good[j] += x[j]
    f = True
    for i in good:
        if i > 1:
            f = False
            break
    if f:
        print("YES")
    else:
        print("NO")
