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.
# 1312C adding powers
import sys

def get_coefs(n, b):
    coefs = []
    while n > 0:
        coefs.insert(0, n % b)
        n  = n // b
    return coefs

t = int(input())

# the trick is to find the coeffeceints in the base k expansion of n 
# i.e find the array a_0, a_1, a_2,... of n = a_0 k^0 + a_1 k^1 +....
# the coeffeceints must either be 0 or 1 otherwise we would have to repeat an operation 
# in addition, no two array elements can agree on the same coeffecient position i.e.
# x&y=0 <=> a_i&b_i = 0 for all i


for _ in range(t):
    n, k = list(map(int, input().split()))
    arr  = list(map(int, input().split()))
    coefs = [] 
    possible  = True
    for num in arr:
        coefs.append(get_coefs(num, k))
    # pad the subarrays with 0s
    # first find largest subarray 
    m = 0
    for x in coefs:
        if len(x) >0:
            m = max(len(x), m)
    # pad
    if m == 0:
        print("YES")
        continue
    else:
        for i in range(n):
            while len(coefs[i]) < m:
                coefs[i].insert(0,0)
                
    # add up subarrays
    sum_arr = [0]*m
    for y in range(m):
        for x in range(n):
            sum_arr[y] += coefs[x][y]
    
    for a in sum_arr:
        if a>1:
            possible = False
            break
    if possible:
        print("YES")
    else: print("NO")