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;
long long int n, k, t;
vector<long long int> a;
vector<long long int> mosla;
void fun(long long int base, long long int p) {
  long long int mul = 1;
  mosla.push_back(1);
  for (long long int i = 1; i <= p; i++) {
    mul *= base;
    mosla.push_back(mul);
  }
  sort(mosla.rbegin(), mosla.rend());
}
map<long long int, long long int> used;
bool isOk(long long int num) {
  for (long long int i = 0; i < mosla.size(); i++) {
    if (num > 0 && mosla[i] <= num && used[mosla[i]] == 0) {
      num -= mosla[i];
      used[mosla[i]]++;
    }
  }
  if (num == 0) return 1;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> k;
    long long int p = (log10(1e16) / log10(k)) + 1;
    fun(k, p);
    for (long long int i = 0; i < n; i++) {
      long long int x;
      cin >> x;
      if (x != 0) a.push_back(x);
    }
    if (a.size() == 0)
      cout << "YES" << endl;
    else {
      sort(a.begin(), a.end());
      bool ck = 1;
      for (long long int i = 0; i < a.size(); i++) {
        if (!isOk(a[i])) ck = 0;
      }
      if (!ck)
        cout << "NO" << endl;
      else
        cout << "YES" << endl;
    }
    mosla.clear();
    used.clear();
    a.clear();
  }
  return 0;
}
