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 examA():
    T = I()
    ans = []
    for _ in range(T):
        N, M = LI()
        if N%M!=0:
            ans.append("NO")
        else:
            ans.append("YES")
    for v in ans:
        print(v)
    return

def examB():
    T = I()
    ans = []
    for _ in range(T):
        N = I()
        A = LI()
        A.sort()
        ans.append(A[::-1])
    for v in ans:
        print(" ".join(map(str,v)))
    return

def examC():
    T = I()
    ans = []
    for _ in range(T):
        N, K = LI()
        A = LI()
        sumA = sum(A)
        if sumA==0:
            ans.append("YES")
            continue
        cur = 0
        L = []
        for i in range(100):
            now = K**i
            L.append(now)
            cur += now
            if cur>=sumA:
                break
        for i in range(N):
            A[i] *= (-1)
        heapify(A)
        #print(A)
        for l in L[::-1]:
            if not A:
                break
            a = -heappop(A)
            if a<l:
                heappush(A, -a)
            elif a>l:
                heappush(A,-(a-l))
        if not A or heappop(A)==0:
            ans.append("YES")
        else:
            ans.append("NO")
    for v in ans:
        print(v)
    return

def examD():
    ans = 0
    print(ans)
    return

def examE():
    ans = 0
    print(ans)
    return

def examF():
    ans = 0
    print(ans)
    return

import sys,copy,bisect,itertools,heapq,math,random
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = 10**(-12)
alphabet = [chr(ord('a') + i) for i in range(26)]

if __name__ == '__main__':
    examC()

"""

"""