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.
from collections import defaultdict
import collections
from heapq import heappush, heappop, heapify
from collections import deque, defaultdict, Counter
import itertools
from itertools import permutations, combinations, accumulate
import sys
import bisect
import string
import math
import time


def I(): return int(input())


def MI(): return map(int, input().split())


def S(): return input()


def MS(): return map(str, input().split())


def LI(): return [int(i) for i in input().split()]


def LI_(): return [int(i)-1 for i in input().split()]


def StoI(): return [ord(i)-97 for i in input()]


def ItoS(nn): return chr(nn+97)


def input(): return sys.stdin.readline().rstrip()


def print_matrix(mat):
    for i in range(len(mat)):
        print(*mat[i])


def show(*inp, end='\n'):
    if show_flg:
        print(*inp, end=end)


YN = {False: 'No', True: 'Yes'}
MOD = 10**9+7
inf = float('inf')
IINF = 10**10
l_alp = string.ascii_lowercase
u_alp = string.ascii_uppercase
nums = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '10']

show_flg = False
# show_flg = True


def solve(T):
    n, k = MI()
    a = LI()
    MAX = 10**16 + 1
    v = 1

    while v < MAX:
        v *= k

    # print(v)

    while v > 0:
        for i in range(n):
            if a[i] >= v:
                a[i] -= v
                break
        v //= k

    # print(*a)
    if all([a[i] == 0 for i in range(n)]):
        print('YES')
    else:
        print('NO')


def main():
    T = I()

    for i in range(T):
        solve(T)


if __name__ == '__main__':
    main()
