#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int M = 1e5 + 3;
const int N = 4;
const int mo = 1e9 + 7;
const LL inf = 1e18 + 1;
map<int, LL> mp;
LL a[M];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    mp[x]++;
  }
  int cnt = 0;
  int sum = 0;
  for (auto &e : mp) {
    a[cnt++] = e.second;
    sum += e.second;
    if (sum >= 3) break;
  }
  LL res = 0;
  if (cnt == 1)
    res = a[0] * (a[0] - 1) * (a[0] - 2) / 6;
  else if (cnt == 2) {
    if (a[0] == 1)
      res = a[1] * (a[1] - 1) / 2;
    else
      res = a[1];
  } else
    res = a[2];
  printf("%I64d\n", res);
  return 0;
}

### Prompt

Generate a Cpp solution to 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;
using LL = long long;
const int M = 1e5 + 3;
const int N = 4;
const int mo = 1e9 + 7;
const LL inf = 1e18 + 1;
map<int, LL> mp;
LL a[M];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    mp[x]++;
  }
  int cnt = 0;
  int sum = 0;
  for (auto &e : mp) {
    a[cnt++] = e.second;
    sum += e.second;
    if (sum >= 3) break;
  }
  LL res = 0;
  if (cnt == 1)
    res = a[0] * (a[0] - 1) * (a[0] - 2) / 6;
  else if (cnt == 2) {
    if (a[0] == 1)
      res = a[1] * (a[1] - 1) / 2;
    else
      res = a[1];
  } else
    res = a[2];
  printf("%I64d\n", res);
  return 0;
}

```