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;
using lli = long long;
using lld = long double;
using ulli = unsigned long long int;
using pll = pair<lli, lli>;
using ttt = pair<lli, pll>;
using vttt = vector<ttt>;
using vll = vector<pll>;
using vl = vector<lli>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using cd = complex<lld>;
void chk(long long int a, int k, vector<int>& B, int& ck) {
  long long int cur = 0;
  int i = 1;
  if (a % k != 0) {
    if ((a - 1) % k != 0) {
      ck = 0;
      return;
    } else {
      a--;
      B.push_back(0);
    }
  }
  while (a > 0) {
    if (pow(k, i) == a) {
      B.push_back(i);
      break;
    } else if (pow(k, i) > a) {
      B.push_back(i - 1);
      a = a - pow(k, i - 1);
      i = 1;
    } else if (pow(k, i) < a) {
      i++;
    }
  }
}
void sanjeev(int kkkk) {
  int t;
  cin >> t;
  while (t > 0) {
    t--;
    int n, k;
    cin >> n >> k;
    vector<long long int> A;
    for (int i = 0; i < n; i++) {
      long long int tmp;
      cin >> tmp;
      A.push_back(tmp);
    }
    vector<int> B;
    int ck = 1;
    for (int i = 0; i < A.size() && ck == 1; i++) {
      if (A[i] == 0) continue;
      chk(A[i], k, B, ck);
    }
    map<int, int> hash;
    for (int i = 0; i < B.size(); i++) {
      long long int k = B[i];
      if (hash.find(k) == hash.end())
        hash[k] = 1;
      else {
        hash[k] = hash[k] + 1;
      }
    }
    vector<int> fre;
    for (auto it = hash.begin(); it != hash.end(); it++) {
      int p = it->second;
      fre.push_back(p);
    }
    for (int i = 0; i < fre.size(); i++)
      if (fre[i] > 1) {
        ck = 0;
        break;
      }
    if (ck == 1)
      cout << "YES";
    else
      cout << "NO";
    cout << '\n';
  }
}
int main() {
  if (!0) ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(10);
  int t = 1;
  for (int i = 0; i < t; i++) {
    sanjeev(i + 1);
  }
  if (0) system("pause");
}
