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.
t=int(input())
for _ in range(t):
	
	n,k=map(int,input().split())
	al=list(map(int,input().split()))
	al2=[]
	for i in range(n):
		if (al[i]!=0):
			al2.append(al[i])

	al=al2	
	al.sort(reverse=True)

	n=len(al)
	if (n==0):
		print("YES")
		continue

	#print("al = ",al)
	x=1
	pl=[]

	while (x<=al[0]):
		pl.append(x)
		x=x*k

	#print("pl = ",pl)

	i=0
	while (i<n):

		j=len(pl)-1

		while (j>=0 and (pl[j]==0 or pl[j]>al[i])):
			j-=1

		if (j<0):
			break
		if (al[i]>0):
			al[i]-=pl[j]
			pl[j]=0

		if (al[i]==0):
			#print("pl = ",pl," al = ",al)
			i+=1

	if (i==n):
		print("YES")
	else:
		print("NO")