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.
def _add_power(arr, base, arr_len):
    max_n = max(arr)
    res = [1]
    while res[-1] * base <= max_n:
        res.append(res[-1] * base)
    def dfs(arr, base, res, visited):
        if not arr: return True

        if arr[0] == 0: return dfs(arr[1:], base, res, visited)
        if arr[0] > sum([res[i] for i in range(len(res)) if not visited[i]]): return False
        
        ans = [False]
        for i in range(len(res)):
            if visited[i] == 1: continue
            if arr[0] == res[i]:
                visited[i] = 1
                ans.append(dfs(arr[1:], base, res, visited))
                break
            if arr[0] > res[i]:
                arr[0] = arr[0]-res[i]
                visited[i] = 1
                ans.append(dfs(arr, base, res, visited))
                visited[i] = 0
                arr[0] = arr[0]+ res[i]
        return any(ans)
        

    arr.sort()
    res.sort(reverse=True)
    #print(res, arr)
    visited = [0] * len(res)
    return dfs(arr, base, res, visited)

def add_power(arr, base, arr_len):
    max_n = max(arr)
    res = [1]
    while res[-1] * base <= max_n:
        res.append(res[-1] * base)

    def dfs(n, res, visited, used):
        if n == 0:
            for i in visited:
                used.append(i)
            #print(used)
            return True
        if n < 0: return False
        if not res or sum(res) < n: return False
        return dfs(n-res[-1], res[:-1], [res[-1]] + visited, used) or dfs(n, res[:-1], visited, used)
        

    arr.sort(reverse=True)
    res.sort()
    for n in arr:
        memo = {}
        used = []
        if not dfs(n, res, [], used):
            return False
        res = [i for i in res if i not in used]
        #print(n, used, res)
    #print(ans)
    return True
        

t = int(input())  # read a line with a single integer
for i in range(1, t + 1):
    arr_len, base = [int(s) for s in input().split(" ")]  # read a list of integers, 2 in this case
    arr = [int(s) for s in input().split(" ")]  # read a list of integers, 2 in this case
    h = add_power(arr, base, arr_len)
    ans = ["NO", "YES"][h]
    print(ans)
