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 sys import stdin
from collections import Counter



T = int(input())
for _ in range(T):
	n, k = map(int, stdin.readline().split())
	a = list(map(int, stdin.readline().split()))

	count = 0
	for i in range(len(a)):
		if a[i] % k == 1:
			a[i] -= 1
			count += 1

	if count > 1:
		print ('NO')
		continue 

	zz = False

	x = Counter()
	y = Counter()

	while set(a) != {0}:
		for i in range(len(a)):
			if a[i] == 0:
				continue 
			if a[i] % k != 0:
				print ('NO')
				zz = True
				break

			else:
				e = 0
				while True:
					if a[i] % k**e == 0:
						e += 1
					else:
						break 
				y[e - 1 + x[i]] += 1
				if y[e - 1 + x[i]] > 1:
					print ('NO')
					zz = True 
					break 
				a[i] = a[i]//k**(e - 1) - 1
				x[i] += e - 1

		if zz:
			break

	if zz:
		continue 
	else:
		print ('YES')





