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.
import os, sys, bisect, copy, math
from collections import defaultdict, Counter, deque
from functools import lru_cache   #use @lru_cache(None)
if os.path.exists('in.txt'): sys.stdin=open('in.txt','r')
if os.path.exists('out.txt'): sys.stdout=open('out.txt', 'w')
#
def input(): return sys.stdin.readline()
def mapi(arg=0): return map(int if arg==0 else str,input().split())
#------------------------------------------------------------------
def deconstruct(aNum, aBase):
    primeComps = []  #initialise array to hold the indexes
    num = aNum
    while num >= 1:
        #index = int(math.floor(math.log(num,aBase)))
        index = ilog(num,aBase)
        num = num - (aBase ** index)
        primeComps.append(int(index)) # append each index into the array
    return primeComps

def ilog(aNum, aBase):
    count = 0
    temp = aNum
    while temp >= (aBase):
        count = count + 1
        temp = temp/aBase
    return count

for _ in range(int(input())):
    n,k = mapi()
    a = list(mapi())
    v = [0]*n
    check = []
    a.sort()
    for i in range(n):
        if a[i]!=0:
            a = a[i:]
            break
    if a.count(0)==len(a):
        print("YES")
        continue
    if len(set(a))!=len(a):
        print("NO")
        continue
    for i in a:
        tmp=deconstruct(i,k)
        check+=tmp
    #print(check)
    if len(set(check))==len(check):
        print("YES")
    else:
        print("NO")
