#include <bits/stdc++.h>
using namespace std;
int const inf = 1000 * 1000 * 1000;
long long const inf64 = 1ll * inf * inf;
int const N = 1e5 + 5;
int n;
int a[N];
int b[N];
int sz;
long long gcd(long long x, long long y) { return x == 0 ? y : gcd(y % x, x); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    long long tmp, x;
    x = a[i];
    tmp = gcd(x, a[1]);
    x /= tmp;
    tmp = gcd(x, a[2]);
    x /= tmp;
    tmp = gcd(x, a[3]);
    x /= tmp;
    if (x == 1) {
      b[sz++] = a[i];
    }
  }
  long long res = 0;
  map<long long, long long> cn;
  map<long long, long long> have;
  long long tmp, A, B, C, x;
  for (int i = 0; i < sz; i++) {
    A = a[1];
    B = a[2];
    C = a[3];
    x = b[i];
    tmp = gcd(x, A);
    x /= tmp;
    A /= tmp;
    tmp = gcd(x, B);
    x /= tmp;
    B /= tmp;
    tmp = gcd(x, C);
    x /= tmp;
    C /= tmp;
    if ((double)A * (double)B * (double)C > 1e18 + 100) {
    } else {
      long long need = A * B * C;
      res += cn[need];
    }
    for (pair<long long, long long> iter : have) {
      cn[iter.first * (long long)b[i]] += iter.second;
    }
    have[b[i]]++;
  }
  cout << res << "\n";
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

After returning from the army Makes received a gift — an array a consisting of n positive integer numbers. He hadn't been solving problems for a long time, so he became interested to answer a particular question: how many triples of indices (i, j, k) (i < j < k), such that ai·aj·ak is minimum possible, are there in the array? Help him with it!

Input

The first line of input contains a positive integer number n (3 ≤ n ≤ 105) — the number of elements in array a. The second line contains n positive integer numbers ai (1 ≤ ai ≤ 109) — the elements of a given array.

Output

Print one number — the quantity of triples (i, j, k) such that i, j and k are pairwise distinct and ai·aj·ak is minimum possible.

Examples

Input

4
1 1 1 1


Output

4


Input

5
1 3 2 3 4


Output

2


Input

6
1 3 3 1 3 2


Output

1

Note

In the first example Makes always chooses three ones out of four, and the number of ways to choose them is 4.

In the second example a triple of numbers (1, 2, 3) is chosen (numbers, not indices). Since there are two ways to choose an element 3, then the answer is 2.

In the third example a triple of numbers (1, 1, 2) is chosen, and there's only one way to choose indices.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int const inf = 1000 * 1000 * 1000;
long long const inf64 = 1ll * inf * inf;
int const N = 1e5 + 5;
int n;
int a[N];
int b[N];
int sz;
long long gcd(long long x, long long y) { return x == 0 ? y : gcd(y % x, x); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    long long tmp, x;
    x = a[i];
    tmp = gcd(x, a[1]);
    x /= tmp;
    tmp = gcd(x, a[2]);
    x /= tmp;
    tmp = gcd(x, a[3]);
    x /= tmp;
    if (x == 1) {
      b[sz++] = a[i];
    }
  }
  long long res = 0;
  map<long long, long long> cn;
  map<long long, long long> have;
  long long tmp, A, B, C, x;
  for (int i = 0; i < sz; i++) {
    A = a[1];
    B = a[2];
    C = a[3];
    x = b[i];
    tmp = gcd(x, A);
    x /= tmp;
    A /= tmp;
    tmp = gcd(x, B);
    x /= tmp;
    B /= tmp;
    tmp = gcd(x, C);
    x /= tmp;
    C /= tmp;
    if ((double)A * (double)B * (double)C > 1e18 + 100) {
    } else {
      long long need = A * B * C;
      res += cn[need];
    }
    for (pair<long long, long long> iter : have) {
      cn[iter.first * (long long)b[i]] += iter.second;
    }
    have[b[i]]++;
  }
  cout << res << "\n";
  return 0;
}

```