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 fac[500005];
map<long long, long long> prime;
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = ((res % p) * (x % p)) % p;
    y = y >> 1;
    x = ((x % p) * (x % p)) % p;
  }
  return res;
}
long long modInverse(long long n, long long p) { return power(n, p - 2, p); }
long long nCrModPFermat(long long n, long long r, long long p) {
  if (r == 0) return 1;
  if (r == n) return 1;
  return (((fac[n] % p) * (modInverse(fac[r], p) % p)) % p *
          (modInverse(fac[n - r], p) % p)) %
         p;
}
void sieve(long long n) {
  for (long long p = 2; p * p <= n; p++) {
    if (!prime[p]) {
      for (long long i = p * p; i <= n; i += p) prime[i] = 1;
    }
  }
}
void fun() {
  long long n, k;
  cin >> n >> k;
  map<long long, long long> mp1, mp2;
  for (long long i = 0; i <= 64; i++) {
    mp1[i] = 1;
  }
  long long arr[n];
  for (long long i = 0; i < n; i++) cin >> arr[i];
  for (long long i = 0; i < n; i++) {
    if (arr[i] == 0) continue;
    long long n1 = arr[i];
    long long ind = 0;
    while (n1) {
      if (n1 % k == 0) {
        n1 = n1 / k;
        ind++;
        continue;
      } else if (n1 % k == 1) {
        if (mp1[ind] > 0) {
          mp1[ind]--;
          ind++;
          n1 = n1 / k;
          continue;
        } else {
          cout << "NO\n";
          return;
        }
      } else {
        cout << "NO\n";
        return;
      }
    }
  }
  cout << "YES\n";
}
signed main() {
  long long tc;
  tc = 1;
  cin >> tc;
  while (tc--) {
    fun();
  }
}
