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;
using ll = long long;
ll nearest(ll x, int base) {
  int ret = 0;
  for (int i = 0; i < 60; i++) {
    if (pow(base, i) > x) {
      ret = i - 1;
      break;
    }
  }
  return ret;
}
bool bad = 0;
vector<int> get(ll x, int base) {
  vector<int> v;
  while (x >= base) {
    int xx = nearest(x, base);
    v.push_back(xx);
    x -= pow(base, xx);
  }
  if (x == 1) {
    v.push_back(0);
    x--;
  }
  if (x != 0) {
    bad = 1;
  }
  return v;
}
int main(int argc, char const *argv[]) {
  int t;
  scanf("%d", &t);
  while (t--) {
    bad = 0;
    int n, base;
    scanf("%d %d", &n, &base);
    vector<ll> a(n);
    vector<int> cnt(61);
    for (int i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
      if (a[i] == 0) continue;
      vector<int> ret = get(a[i], base);
      for (auto j : ret) {
        cnt[j]++;
      }
    }
    if (bad == 1) {
      printf("NO\n");
    } else {
      for (int i = 0; i <= 60; i++) {
        if (cnt[i] > 1) {
          bad = 1;
          break;
        }
      }
      if (bad) {
        printf("NO\n");
      } else {
        printf("YES\n");
      }
    }
  }
  return 0;
}
