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;
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:16777216")
long long linf = 2 * 1000000000000000000;
template <typename T>
void input(vector<T> &a) {
  for (int i = 0; i < a.size(); i++) cin >> a[i];
}
template <typename T>
void print(vector<T> &a) {
  for (auto to : a) cout << to << ' ';
}
vector<int> parent, rang;
void make_set(int v) {
  parent[v] = v;
  rang[v] = 0;
}
int find_set(int v) {
  if (v == parent[v]) return v;
  return parent[v] = find_set(parent[v]);
}
void union_sets(int a, int b) {
  a = find_set(a);
  b = find_set(b);
  if (a != b) {
    if (rang[a] < rang[b]) swap(a, b);
    parent[b] = a;
    if (rang[a] == rang[b]) ++rang[a];
  }
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<int> cnt(60, 0);
  vector<long long> st;
  long long now = 1;
  st.push_back(now);
  while (now * k <= 1e16) {
    now *= k;
    st.push_back(now);
  }
  vector<long long> a(n);
  input(a);
  for (int i = 0; i < n; i++) {
    long long d = a[i];
    for (int i = st.size() - 1; i >= 0; i--) {
      while (d >= st[i]) {
        d -= st[i];
        cnt[i]++;
      }
    }
  }
  for (int i = 0; i < cnt.size(); i++) {
    if (cnt[i] > 1) {
      cout << "NO" << '\n';
      return;
    }
  }
  cout << "YES" << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int k = 1;
  cin >> k;
  while (k--) solve();
}
