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;
long long powerr(long long x, long long y);
long long mpower(long long x, long long y, long long p);
long long gcdExtended(long long a, long long b);
bool isPrime(long long n);
vector<long long> v[100001];
char v1[100001];
long long f = 0;
long long A[100001];
void dividn(long long A1[], long long l, long long r, long long p) {
  if (l > r) {
    return;
  } else {
    long long i;
    long long mx = 0;
    long long z;
    for (i = l; i < r + 1; i++) {
      if (A1[i] > mx) {
        mx = A1[i];
        z = i;
      }
    }
    A[z] = p;
    p++;
    dividn(A1, l, z - 1, p);
    dividn(A1, z + 1, r, p);
  }
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
map<long long, long long> M;
long long pos;
bool check_sum(long long n) {
  if (n == 1) {
    if (M[f] > 0)
      return false;
    else {
      M[f] = 1;
      return true;
    }
  } else if (n % pos == 1) {
    if (M[f] > 0)
      return false;
    else {
      M[f] = 1;
      return check_sum((n - 1));
    }
  } else if (n % pos == 0) {
    f++;
    return check_sum(n / pos);
  } else
    return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t, j, k, l, h, p, q, r, x, i, y, z, a, b, c, d, n, m;
  char ch, ch2, ch3;
  string s1, s2, s3;
  set<pair<long long, long long> > S1, S;
  t = 1;
  cin >> t;
  x = 0;
  long long N = pow(2, 32) - 1;
  while (t--) {
    cin >> n >> pos;
    long long A[n];
    for (i = 0; i < n; i++) cin >> A[i];
    long long flag = 0;
    M.clear();
    for (i = 0; i < n; i++) {
      if (A[i] > 0) {
        f = 0;
        if (check_sum(A[i]) == 0) {
          flag = 1;
          break;
        }
      }
    }
    if (flag)
      cout << "NO";
    else
      cout << "YES";
    cout << "\n";
  }
}
long long gcdExtended(long long a, long long b) {
  if (b == 0) {
    return a;
  } else
    return gcdExtended(b, a % b);
}
long long mpower(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long powerr(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  long long p = sqrt(n);
  for (int i = 5; i <= p; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
