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 n, k, p[100005];
map<vector<int>, int> m;
long long ans;
int main() {
  for (int i = 2; i < 100005; i++)
    if (!p[i]) {
      p[i] = i;
      for (long long j = (long long)i * i; j < 100005; j += i) p[j] = i;
    }
  scanf("%d%d", &n, &k);
  while (n--) {
    int tmp;
    scanf("%d", &tmp);
    vector<pair<int, int>> v;
    while (tmp != 1) {
      int ta = p[tmp], tb = 0;
      while (tmp % ta == 0) {
        tmp /= ta;
        tb++;
      }
      if (tb % k) v.emplace_back(ta, tb % k);
    }
    sort(v.begin(), v.end());
    vector<int> va, vb;
    for (auto& i : v) {
      va.push_back(i.first * 100 + i.second);
      vb.push_back(i.first * 100 + k - i.second);
    }
    ans += m[va];
    m[vb]++;
  }
  printf("%lld\n", ans);
}
