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 fact(x):
	if x == 0:
		return 1
	else:
		return x * fact(x-1)
def lower_bound(li, num): #return 0 if all are greater or equal to
	answer = -1 #no role though
	start = 0
	end = len(li)-1

	while(start <= end):
		middle = (end+start)//2
		if li[middle] >= num:
			answer = middle
			end = middle - 1
		else:
			start = middle + 1
	return answer #index where x is not less than num
def upper_bound(li, num): #return n-1 if all are small or equal
	answer = -1
	start = 0
	end = len(li)-1

	while(start <= end):
		middle = (end+start)//2

		if li[middle] <= num:
			answer = middle
			start = middle + 1
		
		else:
			end = middle - 1
	return answer #index where x is not greater than num

def abs(x):
	return x if x >=0 else -x
def binary_search(li, val, lb, ub): # ITS A BINARY
	ans = 0
	while(lb <= ub):
		mid = (lb+ub)//2
		#print(mid, li[mid])
		if li[mid] > val:
			ub = mid-1
		elif val > li[mid]:
			lb = mid + 1
		else:
			ans = 1
			break
	return ans

def sieve_of_eratosthenes(n):
	ans = []
	arr = [1]*(n+1)
	arr[0],arr[1], i = 0, 0, 2
	while(i*i <= n):
		if arr[i] == 1:
			j = i+i
			while(j <= n):
				arr[j] = 0
				j += i
		i += 1
	for k in range(n):
		if arr[k] == 1:
			ans.append(k)
	return ans
def nc2(x):
	if x == 1:
		return 0
	else:
		return x*(x-1)//2
def kadane(x):
	sum_so_far = 0
	current_sum = 0
	for i in x:
		current_sum += i
		if current_sum < 0:
			current_sum = 0
		else:
			sum_so_far = max(sum_so_far,current_sum)
	return sum_so_far
def mex(li):
	check = [0]*1001
	for i in li:
		check[i] += 1
	for i in range(len(check)):
		if check[i] == 0:
		 return i 
def sumdigits(n):
	ans = 0
	while(n!=0):
		ans += n%10
		n //= 10
	return ans

def product(li):
	ans = 1
	for i in li:
		ans *= i
	return ans
def maxpower(n,k):
	cnt = 0
	while(n>1):
		cnt += 1
		n //= k
	return cnt

#_______________________________________________________________#
'''
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'''
from math import *
from collections import deque
max_steps = 100005
for _ in range(int(input())):
#for _ in range(1):
	#n = int(input())
	n,k = map(int,input().split())
	#st = input()
	a = list(map(int,input().split()))
	#b = list(map(int,input().split()))
	#a = list(input().split())
	p = maxpower(10000000000000000,k)
	f = 1
	while(p>=-1 and f==1):
		if a.count(0) == n:
			#print('here>>>')
			print('yes')
			break
		elif p == -1:
			f = 0
		for i in range(n):
			if a[i] < 0:
				f = 0
		ind = []
		for i in range(n):
			if a[i] >= k**p:
				a[i] -= k**p
				ind.append(i)
			
		#print(ind,p,k**p,f)

		if len(ind) > 1:
			f = 0
			break
		else:
			p -= 1
	if f == 0:
		#print('>>>>>>>>>>>>>>>>>')
		print('no')

	

		
		









	




 