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;
int dcmp(double a, double b) {
  return ((fabs(a - b) < (1e-9)) ? 0 : (a < b) ? -1 : +1);
}
double crossP(const complex<double> &v1, const complex<double> &v2) {
  return (conj(v1) * v2).imag();
}
double dotP(const complex<double> &v1, const complex<double> &v2) {
  return (conj(v1) * v2).real();
}
double lengthSqr(const complex<double> &p) { return dotP(p, p); }
double length(const complex<double> &p) { return hypot(p.real(), p.imag()); }
double fixAngle(double a) { return ((a > 1) ? 1 : (a < -1) ? -1 : a); }
bool same(const complex<double> &p1, const complex<double> &p2) {
  return ((dcmp(p1.real(), p2.real()) == 0 && dcmp(p1.imag(), p2.imag()) == 0)
              ? 1
              : 0);
}
double getAngle_A_abc(double a, double b, double c) {
  return acos(fixAngle((b * b + c * c - a * a) / (2 * b * c)));
}
const int N = 1000 + 9;
long long x[N];
bitset<N> vis;
bool f(const int n, const long long base) {
  for (auto i = 0; i < n; ++i) {
    int p = -1;
    while (x[i]) {
      p++;
      int r = x[i] % base;
      x[i] /= base;
      if (!r) continue;
      if (r > 1 || vis[p]) return 0;
      vis[p] = 1;
    }
  }
  return 1;
}
int main() {
  int tc;
  scanf("%d", &tc);
  while (tc--) {
    int n, b;
    scanf("%d %d", &n, &b);
    for (auto i = 0; i < n; ++i) scanf("%lld", x + i);
    puts(f(n, b) ? "YES" : "NO");
    vis.reset();
  }
  return 0;
}
