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.
#include <bits/stdc++.h>
using namespace std;
void online_judge() {}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  online_judge();
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, k;
    cin >> n >> k;
    long long int a[n];
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
    }
    bool flag = 1;
    set<long long int> s;
    for (long long int i = 0; i < n; ++i) {
      long long int index = 0;
      while (1) {
        if (a[i] == 0) {
          break;
        }
        long long int temp = 1;
        long long int cnt = 0;
        while ((temp * k) <= a[i]) {
          temp *= k;
          cnt++;
        }
        if (s.count(cnt)) {
          flag = 0;
          break;
        }
        s.insert(cnt);
        a[i] -= temp;
      }
    }
    if (flag) {
      cout << "YES\n";
    } else {
      cout << "NO\n";
    }
  }
}
