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;
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) {
      res = (res * x);
    }
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long mcd(long long a, long long b) {
  if (b > a) swap(a, b);
  if (b == 0)
    return a;
  else
    return mcd(b, a % b);
}
long long t, n;
bool esprimo(long long n) {
  long long s = 0;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      s++;
    }
  }
  if (s >= 2) {
    return false;
  }
  return true;
}
int main() {
  cin >> t;
  while (t--) {
    long long k;
    cin >> n >> k;
    long long a[n];
    multiset<long long> b;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long x = 0, j = 0;
    while (x < 10000000000000000) {
      x = power(k, j, 1000000000000000000);
      b.insert(x);
      j++;
    }
    int q = 1;
    multiset<long long> c, d;
    for (int i = 0; i < n; i++) {
      x = a[i];
      long long p = 0, r = 0;
      if (x % k != 0) {
        x--;
        c.insert(1);
      }
      while (x) {
        if (x % k == 0) {
          x /= k;
          p++;
        } else if (p) {
          x--;
          if (x % k == 0) {
            c.insert(power(k, p, 100000000000000000));
          } else {
            r++;
            x = 0;
          }
          p == 0;
        } else {
          r++;
          x = 0;
        }
      }
      if (r) {
        q = 0;
        break;
      }
    }
    if (q) {
      multiset<long long>::iterator it;
      for (it = c.begin(); it != c.end(); it++) {
        if (b.find(*it) != b.end()) {
          b.erase(*it);
        } else {
          q = 0;
          break;
        }
      }
    }
    if (q) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
    b.clear();
    c.clear();
  }
}
