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 math as m

def f1():
	N=int(input())
	Matrix=list()
	for j in range(0,N):
		M=int(input())
		input1=[int(i) for i in input().split(" ")]
		input1.sort(reverse=True)
		for i in range(0,len(input1)):
			print(input1[i],end=" ")
		print("")

def f2():
	N=int(input())
	Matrix=list()
	for i in range(0,N):
		input1=[int(i) for i in input().split(" ")]
		if((input1[0]%input1[1])==0):
			print("YES")
		else:
			print("NO")

def f3():
	Testcases=int(input())
	for i in range(0,Testcases):
		l1=[int(j) for j in input().split(" ")]
		l2=[int(k) for k in input().split(" ")]
		k=l1[1]
		flag=False
		counter1=0
		MAX1=0
		max_index=0
		for i in range(0,len(l2)):
			if(l2[i]>MAX1):
				MAX1=l2[i]
				max_index=i

		MAX1=max(l2)
		if(MAX1==0):
			print("YES")
			continue
		s=m.log(MAX1)/m.log(k)
		s=round(s,1)
		s=m.floor(s)
		#print("THe max power was "+str(s))
		ks=m.pow(k,s)
		ks=int(ks)
		#print("ks is "+str(ks))
		if(ks>l2[max_index]):
			ks=ks//k
		#	print("gotcha")
		#print(str(l2[max_index])+str("this is max index!")+str("this is ks")+str(ks))
		#print("xxx")
		#print(l2[max_index]-ks)
		#print("XXXXX")
		l2[max_index]=l2[max_index]-ks
		#print(str(l2[max_index])+str("this is max index!"))
		ks=ks//k
		#print(l2)
		while(ks>0):
			MAX1=l2[0]
			max_index=0
			for i in range(0,len(l2)):
		#		print("this is ks"+str(ks))
		#		print(l2)
				if(l2[i]>ks):
					l2[i]=l2[i]-ks
					ks=ks//k
					flag=True
				elif(l2[i]==ks):
					l2[i]=0
					ks=ks//k
					flag=True
			
			if(flag==False):
				ks=ks//k
			else:
				flag=False
				##good
		#print("___")
		x=max(l2)
		y=min(l2)
		#print(l2)
		#print(x,y)

		if(x==y and x==0):
			print("YES")
		
		else:
			print("NO")
			
		


		

if __name__=="__main__":
	f3()