#include <bits/stdc++.h>
using namespace std;
map<long long, int> p;
set<long long> q;
set<long long>::iterator it;
int s[10];
int main() {
  int i, j, k, n;
  long long x, ans;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld", &x);
    p[x]++;
    q.insert(x);
  }
  j = 0;
  k = 0;
  for (it = q.begin(); it != q.end() && j < 3; it++) {
    j = j + p[*it];
    s[k++] = p[*it];
  }
  if (k == 1) {
    ans = (long long)s[0] * (s[0] - 1) * (s[0] - 2) / 6;
    printf("%lld\n", ans);
    return 0;
  } else if (k == 2) {
    if (s[0] == 1)
      ans = (long long)s[1] * (s[1] - 1) / 2;
    else if (s[0] == 2)
      ans = s[1];
    printf("%lld\n", ans);
    return 0;
  } else {
    ans = s[2];
    printf("%lld\n", ans);
    return 0;
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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> p;
set<long long> q;
set<long long>::iterator it;
int s[10];
int main() {
  int i, j, k, n;
  long long x, ans;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld", &x);
    p[x]++;
    q.insert(x);
  }
  j = 0;
  k = 0;
  for (it = q.begin(); it != q.end() && j < 3; it++) {
    j = j + p[*it];
    s[k++] = p[*it];
  }
  if (k == 1) {
    ans = (long long)s[0] * (s[0] - 1) * (s[0] - 2) / 6;
    printf("%lld\n", ans);
    return 0;
  } else if (k == 2) {
    if (s[0] == 1)
      ans = (long long)s[1] * (s[1] - 1) / 2;
    else if (s[0] == 2)
      ans = s[1];
    printf("%lld\n", ans);
    return 0;
  } else {
    ans = s[2];
    printf("%lld\n", ans);
    return 0;
  }
  return 0;
}

```