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;
const int OO = 0x3f3f3f3f, NegOO = -1 * OO, maxarr = 2e5 + 5, mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0), ios::sync_with_stdio(0), cin.tie(0),
      cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long arr[n];
    map<long long, bool> mp;
    for (int c = 0; c < n; c++) cin >> arr[c];
    bool tmam = 1;
    for (int c = 0; c < n; c++) {
      long long powy;
      while (arr[c]) {
        if (arr[c] < k && arr[c] != 1) {
          tmam = 0;
          break;
        }
        for (int i = 0; i < 60; i++) {
          if (powl(k, i) <= arr[c])
            powy = i;
          else
            break;
        }
        if (!mp[powy])
          arr[c] -= powl(k, powy), mp[powy] = 1;
        else {
          tmam = 0;
          break;
        }
      }
      if (!tmam) break;
    }
    if (tmam)
      cout << "YES"
           << "\n";
    else
      cout << "NO"
           << "\n";
  }
  return 0;
}
