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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
vector<long long int> adj[40];
pair<bool, vector<long long int> > solve(long long int n, long long int k) {
  long long int count = 0;
  vector<long long int> ans, b(65);
  while (k) {
    if (k % n == 0) {
      k /= n;
      count += 1;
    } else if (k % n == 1) {
      k -= 1;
      b[count] += 1;
      if (b[count] > 1) return {false, ans};
    } else
      return {false, ans};
  }
  for (long long int i = 0; i < 62; i++) {
    if (b[i]) ans.push_back(i);
  }
  return {true, ans};
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long int n, k, i;
    cin >> n >> k;
    vector<long long int> arr(n);
    bool ok = true;
    for (i = 0; i < n; i++) cin >> arr[i];
    for (i = 0; i < n; i++) {
      pair<bool, vector<long long int> > temp = solve(k, arr[i]);
      if (!temp.first) {
        ok = false;
        break;
      }
      adj[i] = temp.second;
    }
    if (!ok)
      cout << "NO" << endl;
    else {
      set<long long int> s;
      long long int count = 0;
      for (i = 0; i < n; i++) {
        count += adj[i].size();
        for (auto it : adj[i]) s.insert(it);
      }
      if (s.size() != count)
        cout << "NO" << endl;
      else
        cout << "YES" << endl;
    }
  }
  return 0;
}
