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;
using ii = pair<int, int>;
using i3 = pair<int, ii>;
using li = pair<ll, int>;
using lii = pair<ll, ii>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vii = vector<ii>;
using vli = vector<li>;
using vpll = vector<pll>;
using vi3 = vector<i3>;
using vlii = vector<lii>;
const int N = 2e5 + 5;
const ll INF = 1e17 + 7;
const double eps = 1e-9, PI = acos(-1);
int n, k;
void solve(int testCase) {
  scanf("%d %d", &n, &k);
  vl val;
  ll mul = 1;
  val.push_back(1);
  for (int i = 1;; i++) {
    mul *= 1ll * k;
    if (mul > ll(1e16)) break;
    val.push_back(mul);
  }
  int sz = val.size();
  vector<bool> vis(sz + 1, 0);
  bool sol = 1;
  for (int i = 0; i < n; i++) {
    ll x;
    scanf("%lld", &x);
    if (x == 0) continue;
    for (int j = sz - 1; j >= 0; j--) {
      if (!vis[j] && (val[j] <= x)) {
        x -= val[j];
        vis[j] = 1;
      }
    }
    if (x != 0) sol = 0;
  }
  printf("%s\n", sol ? "YES" : "NO");
}
int main() {
  int t = 1;
  scanf("%d", &t);
  for (int testCase = 1; testCase <= t; testCase++) {
    solve(testCase);
  }
  return 0;
}
