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 long long sz = (long long)1e4 + 5;
const long long mx = (long long)2e5 + 5;
const long long inf = (long long)1e18;
const long long mod = (long long)1e9 + 7;
long long pw(long long x, unsigned long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return pw(x, y / 2) * pw(x, y / 2);
  else
    return x * pw(x, y / 2) * pw(x, y / 2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    map<long long, long long> mp;
    long long n, k, f = 0, cnt;
    cin >> n >> k;
    vector<long long> v(n);
    for (long long i = 0; i < n; i++) {
      cin >> v[i];
    }
    vector<vector<long long>> ad(n, vector<long long>(64, 0));
    for (long long i = 0; i < n; i++) {
      long long m = v[i], j = 0;
      while (m) {
        if (m % k > 0) mp[j]++;
        ad[i][j++] = m % k;
        m /= k;
      }
    }
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < 64; j++) {
        if (ad[i][j] > 1 or mp[j] > 1) {
          f = 1;
          break;
        }
      }
      if (f) break;
    }
    if (f)
      cout << "NO\n";
    else
      cout << "YES\n";
  }
  return 0;
}
