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

for t in range(int(input())):
  #n = int(input())
  n,k = [int(i) for i in input().split(' ')]
  a = [int(i) for i in input().split(' ')]
  a.sort()
  i = 0
  d = False

  """if max(a) == 0:
    print('YES')
    continue
  for i in a:
    if i == 0:
      continue
    top = math.log(i,k)
    if top<=i:
      print('NO')
      d = True 
      break
    else:
      i = floor(math.log(a,k))"""


  if max(a) == 0:
    print('YES')
    continue
  top = math.log(max(a))/math.log(k)
  while i<top+2:
    #print(a,i,k**i)
    for j in range(n):
      #print(a[j]%(k**i),end=' ')
      #print(a[j]%(k**(i+1)),end = ' - ')
      if a[j]%(k**i)!=0:
        print('NO')
        d = True
        break 
      elif a[j]%(k**(i+1))!=0: #and (a[j]-k**i)%(k**(i+1))!=0:
        a[j]-=k**i
        break
    #print()
    i+=1
    if d:
      break
  if not d:
    print('YES')