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 tok(k, num):
    res = []
    while num:
        if num % k == 0:
            res.append(0)
        elif num % k == 1:
            res.append(1)
        else:
            return -1
        num /= k
    u = 0
    for num in res[::-1]:
        u = u * 2 + num
    return u

assert tok(9, 810) == 4 + 8
assert tok(2, 1) == 1
assert tok(2, 3) == 3
assert tok(2, 100) == 100

def solve(n, k, ns):
    st = set()
    ms = []
    for num in ns:
        if num == 0:
            continue
        u = tok(k, num)
        if u == -1:
            return False
        ms.append(u)
    m = len(ms)
    for i in xrange(m):
        for j in xrange(i + 1, m):
            if (ms[i] & ms[j]) != 0:
                return False
    return True

assert solve(4, 100, [0, 0, 0, 0])

T = int(raw_input())
for case_ in xrange(T):
    (n, k) = map(int, raw_input().split())
    ns = map(int, raw_input().split())
    print 'YES' if solve(n, k, ns) else 'NO'
