#include <bits/stdc++.h>
using namespace std;
map<long long int, long long int> mp, mp2;
long long int a[1000005];
long long int ncr(long long int n, long long int r) {
  r = min(n - r, r);
  if (n < r) return 0;
  long long int ans = 1;
  for (long long int i = 0; i < r; i++) ans = ans * (n - i), ans /= (i + 1);
  return ans;
}
int main() {
  long long int n;
  scanf("%lld", &n);
  for (long long int i = 1; i <= (long long int)n; i++)
    scanf("%lld", a + i), mp[a[i]]++;
  sort(a + 1, a + n + 1);
  mp2[a[1]]++;
  mp2[a[2]]++;
  mp2[a[3]]++;
  long long int ans = 1;
  for (auto k : mp2) ans = ans * ncr(mp[k.first], k.second);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
map<long long int, long long int> mp, mp2;
long long int a[1000005];
long long int ncr(long long int n, long long int r) {
  r = min(n - r, r);
  if (n < r) return 0;
  long long int ans = 1;
  for (long long int i = 0; i < r; i++) ans = ans * (n - i), ans /= (i + 1);
  return ans;
}
int main() {
  long long int n;
  scanf("%lld", &n);
  for (long long int i = 1; i <= (long long int)n; i++)
    scanf("%lld", a + i), mp[a[i]]++;
  sort(a + 1, a + n + 1);
  mp2[a[1]]++;
  mp2[a[2]]++;
  mp2[a[3]]++;
  long long int ans = 1;
  for (auto k : mp2) ans = ans * ncr(mp[k.first], k.second);
  printf("%lld\n", ans);
  return 0;
}

```