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;
void faster() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
const long long MAXN = 6e5 + 1;
const long double EPS = 1e-12;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long fact(int n) {
  long long f = 1;
  if ((n == 0) || (n == 1))
    f = 1;
  else
    for (int i = 1; i <= n; i++) f *= i;
  return f;
}
int fib(int n) {
  int a = 0, b = 1;
  for (int i = 0; i < n; ++i) {
    b = b + a;
    a = b - a;
  }
  return b;
}
bool compare(const string &c1, const string &c2) {
  return c1.size() < c2.size();
}
string alph = "abcdefghijklmnopqrstuvwxyz";
int t, n, k, ans = 1;
void solve() {
  cin >> n >> k;
  vector<long long> a(n), fact, index;
  ans = 1;
  for (int i = 0; i < n; ++i) {
    fact.clear();
    cin >> a[i];
    if (a[i] != 0) {
      while (a[i]) {
        fact.push_back(a[i] % k);
        a[i] /= k;
      }
      for (int j = 0; j < fact.size(); ++j) {
        if (fact[j] > 1) {
          ans = 0;
          break;
        }
        if (fact[j] == 1) index.push_back(j);
      }
    }
  }
  if (index.size() != 0) {
    sort(index.begin(), index.end());
    for (int i = 1; i < index.size(); ++i) {
      if (index[i] == index[i - 1]) {
        ans = 0;
        break;
      }
    }
  }
  cout << (ans == 1 ? "YES" : "NO");
}
int main() {
  faster();
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
    cout << "\n";
  }
  return 0;
}
