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 math import *
# from sys import stdin,stdout

def binarySearch(arr,x,i):
	l=i
	r=len(arr)-1 
	while l <= r: 
		mid = (l + r)//2; 
		if arr[mid] == x: 
			return mid 
		elif arr[mid] < x: 
			l = mid + 1
		else: 
			r = mid - 1
	return -1

def js(arr,x):
	l=0
	r=len(arr)-1
	ans=-1
	while(l<=r):
		m=(l+r)//2
		if(arr[m]<=x):
			ans=m
			l=m+1
		else:
			r=m-1
	return ans

def jg(arr,x):
	l=0
	r=len(arr)-1
	ans=-1
	while(l<=r):
		m=(l+r)//2
		if(arr[m]>=x):
			ans=m
			r=m-1
		else:
			l=m+1
	return ans

# def ceil(a,b):
# 	if a%b == 0:
# 		return int(a/b)
# 	else:
# 		return (a//b + 1)

def c(n,k):
	i=64
	while(i>=0):
		if(n%(k**i)==0):
			return i
		i-=1
	return -1



for __ in range(int(input())):
	n,k=map(int,input().split())
	ar=list(map(int,input().split()))
	a=[0]*n
	if(ar==a):
		print("YES")
	else:
		ans="YES"
		d=[0]*65
		for i in range(n):
			nm=ar[i]
			while(nm>0):
				if(nm%k==0):
					t=c(nm,k)
					if(d[t]==1 or t==-1):
						ans="NO"
						break
					else:
						nm-=(k**t)
						d[t]=1
				else:
					if(d[0]==1):
						ans="NO"
						break
					else:
						d[0]=1
						nm-=1
		print(ans)
