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 mod = 1e16 + 7;
const long long oo = 1e9 + 10;
const int Max = 1e6 + 10;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long a[Max];
long long f[Max];
long long n, k;
bool k_(long long n, long long k) {
  int cnt = 0;
  while (n) {
    if (n % k != 0) {
      if (n % k > 1) return false;
      if (f[cnt]) return false;
      f[cnt] = 1;
    }
    n /= k;
    cnt++;
  }
  return true;
}
void solve() {
  cin >> n >> k;
  memset(f, 0, sizeof(f));
  int flag = 1;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (flag && a[i] && !k_(a[i], k)) {
      flag = 0;
    }
  }
  if (flag) {
    cout << "YES\n";
  } else {
    cout << "NO\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int ___T;
  cin >> ___T;
  for (int cs = 1; cs <= ___T; cs++) {
    solve();
  }
  return 0;
}
