You are given n positive integers a_1, …, a_n, and an integer k ≥ 2. Count the number of pairs i, j such that 1 ≤ i < j ≤ n, and there exists an integer x such that a_i ⋅ a_j = x^k.

Input

The first line contains two integers n and k (2 ≤ n ≤ 10^5, 2 ≤ k ≤ 100).

The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^5).

Output

Print a single integer — the number of suitable pairs.

Example

Input


6 3
1 3 9 8 24 1


Output


5

Note

In the sample case, the suitable pairs are:

  * a_1 ⋅ a_4 = 8 = 2^3;
  * a_1 ⋅ a_6 = 1 = 1^3;
  * a_2 ⋅ a_3 = 27 = 3^3;
  * a_3 ⋅ a_5 = 216 = 6^3;
  * a_4 ⋅ a_6 = 8 = 2^3.
#include <bits/stdc++.h>
using namespace std;
int mark[100005];
void build_seive() {
  for (int i = 1; i <= 100000; i++) mark[i] = i;
  for (int i = 2; i <= 100000; i += 2) mark[i] = 2;
  for (int i = 3; i * i <= 100000; i++)
    if (mark[i] == i)
      for (int j = i * i; j <= 100000; j += i)
        if (mark[j] == j) mark[j] = i;
}
int M[100005];
vector<pair<int, int> > f(int val) {
  vector<pair<int, int> > arr;
  int cnt = 0, pre = -1;
  while (val > 1) {
    if (pre != -1 && mark[val] != pre) {
      arr.push_back({pre, cnt});
      cnt = 0;
    }
    pre = mark[val];
    cnt++;
    val /= mark[val];
  }
  if (pre != -1) arr.push_back({pre, cnt});
  return arr;
}
int a[100005];
int add(vector<pair<int, int> > a, int k) {
  long long val = 1;
  for (pair<int, int> u : a) {
    int num = (k - u.second % k) % k;
    for (int j = 1; j <= num; j++) {
      if (1LL * val * u.first > 1e5) return -1;
      val *= u.first;
    }
  }
  return val;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  build_seive();
  int n, k;
  cin >> n >> k;
  long long res = 0;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) {
    int x = a[i];
    vector<pair<int, int> > a = f(x);
    int x_new = 1;
    for (pair<int, int> u : a) {
      int num = u.second % k;
      for (int j = 1; j <= num; j++) x_new *= u.first;
    }
    int val = add(a, k);
    if (val != -1) res += M[val];
    M[x_new]++;
  }
  cout << res;
}
