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>
const int MAXN = 3e5 + 10;
const double eps = 1e-8;
using namespace std;
struct edge {
  int t, v;
  edge *next;
} e[MAXN << 1], *h[MAXN], *o = e;
void add(int x, int y, int vul) {
  o->t = y;
  o->v = vul;
  o->next = h[x];
  h[x] = o++;
}
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
int _, n, k;
long long a[MAXN];
bool vis[105];
int main() {
  _ = read();
  while (_--) {
    for (int i = 0; i <= 104; i++) vis[i] = 0;
    n = read();
    k = read();
    for (int i = 1; i <= n; i++) a[i] = read();
    bool flag = 0;
    for (int i = 1; i <= n; i++) {
      long long x = a[i];
      int num = 0;
      while (x) {
        long long t1 = x % k;
        x -= t1;
        x /= k;
        if (t1 >= 2) {
          flag = 1;
          break;
        }
        if (t1 && vis[num]) {
          flag = 1;
          break;
        } else if (t1)
          vis[num] = 1;
        num++;
      }
      if (flag) break;
    }
    if (flag)
      puts("NO");
    else
      puts("YES");
  }
  return 0;
}
