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>
#pragma GCC optimize("-O3")
using namespace std;
long long MOD = 998244353;
long long mpow(long long a, long long b) {
  if (b >= (MOD - 1)) b %= (MOD - 1);
  if (b == 0) return 1;
  long long t1 = mpow(a, b / 2);
  t1 *= t1;
  t1 %= MOD;
  if (b % 2) t1 *= a;
  t1 %= MOD;
  return t1;
}
long long mpow(long long a, long long b, long long p) {
  if (b == 0) return 1;
  long long t1 = mpow(a, b / 2, p);
  t1 *= t1;
  t1 %= p;
  if (b % 2) t1 *= a;
  t1 %= p;
  return t1;
}
long long modinverse(long long a, long long m) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
long long range(long long l, long long r) { return l + rand() % (r - l + 1); }
pair<long long, long long> hashFn(vector<long long> vc) {
  long long p1 = 31, p2 = 97;
  long long m1 = 1e9 + 9, m2 = 998244353;
  long long hsv1 = 0, hsv2 = 0, p_pow1 = 1, p_pow2 = 1;
  for (long long i = 0; i < vc.size(); i++) {
    hsv1 = ((hsv1 + vc[i] + 1) * p_pow1) % m1;
    p_pow1 = (p_pow1 * p1) % m1;
    hsv2 = ((hsv2 + vc[i] + 1) * p_pow2) % m2;
    p_pow2 = (p_pow2 * p2) % m2;
  }
  return make_pair(hsv1, hsv2);
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> ar(n);
  for (long long i = 0; i < n; i++) {
    cin >> ar[i];
  }
  vector<vector<long long> > dcp(n);
  bool flag = 1;
  for (long long i = 0; i < n; i++) {
    long long pw = 0;
    while (ar[i] != 0) {
      while (ar[i] % k == 0) {
        ar[i] /= k;
        pw++;
      }
      ar[i] -= 1;
      dcp[i].push_back(pw);
      if (ar[i] % k != 0) {
        flag = 0;
        break;
      }
    }
  }
  if (flag == 0) {
    cout << "NO"
         << "\n";
  } else {
    bool flg = 1;
    vector<long long> tpp(80, -1);
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < dcp[i].size(); j++) {
        if (tpp[dcp[i][j]] != -1) {
          flg = 0;
        } else {
          tpp[dcp[i][j]] = i;
        }
      }
    }
    if (flg) {
      cout << "YES"
           << "\n";
    } else {
      cout << "NO"
           << "\n";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    solve();
  }
  return 0;
}
