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.
# Author: S Mahesh Raju
# Username: maheshraju2020
# Date: 29/04/2020

from sys import stdin,stdout
from math import gcd, ceil, sqrt
ii1 = lambda: int(stdin.readline().strip())
is1 = lambda: stdin.readline().strip()
iia = lambda: list(map(int, stdin.readline().strip().split()))
isa = lambda: stdin.readline().strip().split()
mod = 1000000007

tc = ii1()
for i in range(tc):
    n, k = iia()
    nums = [0]
    for i in range(55):
        if nums[-1] >= 10**16:
            break
        nums.append(k ** i)

    arr = iia()
    powers = []
    for i in arr:
        if i > 0:
            cur = 0
            for j in range(55):
                if cur + nums[j] >= i:
                    cur += nums[j]
                    break
                cur += nums[j]

            temp = [j]
            for k in range(j)[::-1]:
                if cur - nums[k] >= i:
                    cur -= nums[k]
                else:
                    temp.append(k)

            if cur != i:
                print("NO")
                break
            powers.extend(temp)
    else:
        if len(powers) == len(set(powers)):
            print("YES")
        else:
            print("NO")


    