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.
#!/usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
from bisect import bisect_left, bisect_right
import sys, itertools, math
sys.setrecursionlimit(10**5)
input = sys.stdin.readline
sqrt = math.sqrt
def LI(): return list(map(int, input().split()))
def LF(): return list(map(float, input().split()))
def LI_(): return list(map(lambda x: int(x)-1, input().split()))
def II(): return int(input())
def IF(): return float(input())
def S(): return input().rstrip()
def LS(): return S().split()
def IR(n):
    res = [None] * n
    for i in range(n):
        res[i] = II()
    return res
def LIR(n):
    res = [None] * n
    for i in range(n):
        res[i] = LI()
    return res
def FR(n):
    res = [None] * n
    for i in range(n):
        res[i] = IF()
    return res
def LIR(n):
    res = [None] * n
    for i in range(n):
        res[i] = IF()
    return res
def LIR_(n):
    res = [None] * n
    for i in range(n):
        res[i] = LI_()
    return res
def SR(n):
    res = [None] * n
    for i in range(n):
        res[i] = S()
    return res
def LSR(n):
    res = [None] * n
    for i in range(n):
        res[i] = LS()
    return res
mod = 1000000007
inf = float('INF')

#solve
def solve():
    t = II()
    for _ in range(t):
        n, k = LI()
        a = LI()
        h = []
        for i in a:
            heappush(h, -i)
        x = 1
        m = max(a)
        while x < m:
            x *= k
        while x >= 1:
            m = -heappop(h)
            if m >= x:
                m -= x
            heappush(h, -m)
            x //= k
        if heappop(h) == 0:
            print("YES")
        else:
            print("NO")
    return


#main
if __name__ == '__main__':
    solve()
