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;
void Fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cerr.tie(NULL);
}
inline int D() {
  int t;
  scanf("%d", &t);
  return t;
}
inline long long llD() {
  long long t;
  scanf("%lld", &t);
  return t;
}
long long MOD = 1e9 + 7;
const int N = 1e2 + 5;
long long power[N];
int OO = 1e6;
long long n, m, k, cnt, SU, idx, len, MN;
long long l, r, mid;
long long a, b, c, d;
vector<int> vec;
int BS(long long n) {
  l = 0, r = len - 1, mid;
  while (l < r) {
    mid = (l + r + 1) >> 1;
    if (power[mid] <= n)
      l = mid;
    else
      r = mid - 1;
  }
  return l;
}
void Gen() {
  len = 0, MN = 1e16 + 5;
  power[len] = 1;
  while (power[len] <= MN) {
    power[len + 1] = power[len] * m;
    ++len;
  }
}
int main() {
  int T = D();
  while (T--) {
    scanf("%lld%lld", &n, &m);
    Gen();
    map<int, int> mp;
    bool flage = true;
    while (n--) {
      scanf("%lld", &cnt);
      while (cnt) {
        idx = BS(cnt);
        cnt -= power[idx];
        ++mp[idx];
        flage &= (mp[idx] == 1);
      }
    }
    puts(flage ? "YES" : "NO");
  }
}
