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 double eps = (long double)1 / 1e15;
long long inf = 1e18, mod2 = 998244353, mod1 = 1e9 + 7;
long long sqr(long long a) { return a * a; }
long long qb(long long a) { return a * a * a; }
long long gcd(long long a, long long b) { return !a ? b : gcd(b % a, a); }
long long binpow(long long a, long long b, long long mod) {
  return b ? (b % 2 ? (a * (sqr(binpow(a, b / 2, mod)) % mod)) % mod
                    : sqr(binpow(a, b / 2, mod)) % mod)
           : 1;
}
long long a[31], bl[55], st[55];
int main() {
  cout.tie(0), cin.tie(0), ios_base::sync_with_stdio(0);
  long long q, n, k, i, j;
  cin >> q;
  while (q--) {
    cin >> n >> k;
    for (i = 1; i <= n; i++) cin >> a[i];
    for (i = 0; i <= 54; i++) bl[i] = st[i] = 0;
    st[0] = 1;
    for (i = 1; i <= 54; i++) st[i] = st[i - 1] * k;
    bool e = 0;
    for (i = 1; i <= n; i++) {
      if (!a[i]) continue;
      while (a[i] > 0) {
        for (j = 1; j < 55; j++)
          if (a[i] % st[j]) break;
        if (bl[j - 1]) e = 1;
        bl[j - 1] = 1;
        a[i] -= st[j - 1];
      }
    }
    cout << (!e ? "YES\n" : "NO\n");
  }
  return 0;
}
