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;
const long long mod = 1e9 + 7;
void debug(string s) { cout << s << "\n"; }
const int MAX = 2000007;
bool prime[MAX];
int spf[MAX];
void sieve() {
  for (int i = 2; i < MAX; i += 2) spf[i] = 2;
  for (long long i = 3; i < MAX; i += 2) {
    if (!prime[i]) {
      spf[i] = i;
      for (long long j = i; (j * i) < MAX; j += 2) {
        if (!prime[j * i]) {
          prime[j * i] = true;
          spf[j * i] = i;
        }
      }
    }
  }
}
long long GCD(long long a, long long b) {
  long long c = a % b;
  while (c != 0) {
    a = b;
    b = c;
    c = a % b;
  }
  return b;
}
long long power(long long a, long long b) {
  long long ans = 1;
  while (b != 0) {
    if (b % 2 == 1) ans *= a;
    a *= a;
    b = b >> 1;
  }
  return ans;
}
void void_main() {
  long long n, k;
  cin >> n >> k;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
  vector<long long> h(100, 0);
  for (long long i = 0; i < n; i++) {
    long long x = v[i];
    long long j = 0;
    while (x) {
      long long rem = x % k;
      h[j++] += rem;
      x /= k;
    }
  }
  bool say = 1;
  for (long long i = 0; i < 100; i++) {
    if (h[i] > 1) {
      say = 0;
      break;
    }
  }
  if (say) {
    cout << "YES\n";
  } else {
    cout << "NO\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int _ = 1;
  cin >> _;
  while (_--) {
    void_main();
  }
  return 0;
}
