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 isValid(long long int num, vector<double> &power) {
  for (int i = power.size() - 1; i >= 0; i--) {
    if (num >= power[i] && power[i] != -1) {
      num -= power[i];
      power[i] = -1;
    }
  }
  if (num == 0) return 1;
  return 0;
}
int main() {
  long long int test = 1;
  cin >> test;
  for (int loop = 0; loop < test; loop++) {
    long long int num, val;
    cin >> num >> val;
    vector<long long int> arr(num);
    for (int i = 0; i < num; i++) cin >> arr[i];
    ;
    sort(arr.begin(), arr.end());
    long long int res = 1;
    vector<long long int> power;
    long long int value = 1;
    power.push_back(1);
    for (int i = 1; i < 100; i++) {
      value = value * val;
      power.push_back(value);
      if (value > 10000000000000000 || value < 0) break;
    }
    for (int i = num - 1; i >= 0; i--) {
      if (arr[i] == 0)
        break;
      else {
        long long int temp = arr[i];
        for (int j = power.size() - 1; j >= 0; j--) {
          if (power[j] > 0 && power[j] <= temp) {
            temp = temp - power[j];
            power[j] = -1;
          }
        }
        if (temp == 0)
          continue;
        else {
          res = 0;
          break;
        }
      }
    }
    if (res)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
}
