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 is_normal(x, q, k, chisl):
    answer = chisl
    f = True
    w = 0
    while f and x > 0:
        f = False
        for i in range(1, q):
            st = k ** i
            if x % st == 0:
                kol = 0
                while x % st == 0:
                    x //= st
                    kol += 1
                x -= 1
                q -= i * kol
                answer.append(i * kol + w)
                w += i * kol
                f = True
                break
    if f:
        return answer
    return 'o'


t = int(input())

for i in range(t):
    n, k = map(int, input().split())
    arr = [int(x) for x in input().split()]
    Max = 0
    d = dict()
    flag = True
    for j in range(n):
        e = arr[j]
        ch = 0
        while e > 1:
            e //= k
            ch += 1
        Max = max(Max, ch + 1)
    
    for j in range(Max):
        d[j] = False
    
    for j in range(n):
        if arr[j] == 0:
            continue
        mass1 = is_normal(arr[j] - 1, Max, k, [0])
        mass2 = is_normal(arr[j], Max, k, [])
        if mass1 == 'o' and mass2 == 'o':
            flag = False
            break
        if mass1 != 'o':
            mass = mass1
        else:
            mass = mass2
        for l in mass:
            if d[l]:
                flag = False
                break
            d[l] = True
            
    if not flag:
        print('NO')
    else:
        print('YES')