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 findlog(long long int num, long long int k) {
  long long int c = 0;
  while (num > 1) {
    num /= k;
    c += 1;
  }
  return c;
}
int main() {
  long long int t, n, k, temp;
  vector<long long int> v;
  cin >> t;
  long long int x;
  set<long long int> s;
  while (t-- > 0) {
    long long int c = 0;
    cin >> n >> k;
    bool flag = true;
    bool f = true;
    long long int maxi = 0;
    for (int i = 0; i < n; i++) {
      cin >> x;
      v.push_back(x);
      if (maxi < x) maxi = x;
      if (x % k != 0 && x % k != 1) flag = false;
    }
    sort(v.begin(), v.end());
    long long int res;
    if (!flag) goto label;
    res = findlog(maxi, k);
    for (int i = 0; i <= res; i++) s.insert(i);
    for (int i = 0; i < n; i++) {
      if (v[i] != 0) {
        while (v[i] != 0) {
          temp = findlog(v[i], k);
          if (find(s.begin(), s.end(), temp) != s.end()) {
            v[i] -= pow(k, temp);
            s.erase(temp);
          } else {
            f = false;
            break;
          }
        }
      }
      if (!f) goto label;
    }
  label:;
    for (int i = 0; i < n; i++) {
      if (v[i] == 0) c++;
    }
    if (c == n)
      cout << "YES\n";
    else
      cout << "NO\n";
    v.clear();
    s.clear();
  }
  return 0;
}
