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;
const int maxn = 100;
long long n, m, T, k;
string s, ss;
long long a[maxn], b[maxn];
long long num[maxn];
map<long long, long long> mp;
struct edge {
  long long nxt, to;
} e[maxn * 2];
struct QQ {
  long long sum, lcol, rcol;
};
long long read() {
  char c = getchar();
  long long x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
bool cmp(long long a, long long b) { return a > b; }
void check(long long x) {
  for (int i = 60; i >= 0; i--) {
    if (num[i] == 0) continue;
    a[i] += x / num[i];
    x %= num[i];
  }
}
signed main() {
  cin >> T;
  long long book, cnt;
  while (T--) {
    book = 0;
    memset(num, 0, sizeof(num));
    memset(a, 0, sizeof(a));
    num[0] = 1;
    cin >> n >> k;
    for (int i = 1; i <= 60; i++) {
      num[i] = num[i - 1] * k;
      if (num[i] > 10000000000000009) break;
    }
    for (int i = 1; i <= n; i++) {
      cin >> m;
      check(m);
    }
    for (int i = 0; i <= 60; i++) {
      if (a[i] > 1) book = 1;
    }
    if (book)
      puts("NO");
    else
      puts("YES");
  }
}
