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;
int main() {
  int t;
  std::cin >> t;
  while (t--) {
    int n, k;
    std::cin >> n >> k;
    std::vector<long long int> a;
    long long int temp;
    for (int i = 0; i < n; i++) {
      std::cin >> temp;
      a.push_back(temp);
    }
    sort(a.begin(), a.end());
    for (int i = 0; i < a.size(); i++) {
      if (a[i] == 0) {
        a.erase(a.begin() + i);
        i--;
      } else {
        break;
      }
    }
    if (a.size() == 0) {
      std::cout << "YES" << endl;
      continue;
    }
    int flag1 = 0;
    for (int i = 1; i < a.size(); i++) {
      if (a[i] == a[i - 1]) {
        flag1 = 1;
        break;
      }
    }
    if (flag1 == 1) {
      std::cout << "NO" << endl;
      continue;
    }
    long long int b = pow(k, 0);
    temp = a[n - 1];
    int count = 0;
    while (temp / b > 1) {
      count++;
      b = pow(k, count);
    }
    int j;
    int flag = 0;
    for (int i = count; i >= 0; i--) {
      j = a.size() - 1;
      int ans = 0;
      int flag2 = 0;
      long long int c = pow(k, i);
      long long int b = a[j] / c;
      while (b == 1) {
        flag2 = 1;
        a[j] = a[j] - c;
        if (j > 0) j--;
        ans++;
        b = a[j] / c;
      }
      if (flag2 == 0) {
        if (a[j] / pow(k, i) > 1) {
          flag = 1;
          break;
        }
      }
      if (ans > 1) {
        flag = 1;
        break;
      }
      sort(a.begin(), a.end());
    }
    if (flag == 0) {
      std::cout << "YES" << endl;
    } else {
      std::cout << "NO" << endl;
    }
  }
}
