#include <bits/stdc++.h>
using namespace std;
int pronadjiPrviIndeks(int niz[3]) {
  for (int i = 0; i < 3; i++) {
    if (niz[i] == -1) return i;
  }
  return -1;
}
long long nCr(long long n, long long k) {
  long long proizvod = 1LL;
  for (int i = 0; i < k; i++) {
    proizvod *= n;
    --n;
  }
  for (int i = 1; i <= k; i++) {
    proizvod /= (long long)i;
  }
  return proizvod;
}
int main() {
  int n;
  cin >> n;
  map<int, int> m;
  for (int i = 0; i < n; i++) {
    int broj;
    cin >> broj;
    ++m[broj];
  }
  int niz[3] = {-1, -1, -1};
  for (map<int, int>::iterator it = m.begin(); it != m.end(); ++it) {
    int indeks = pronadjiPrviIndeks(niz);
    if (indeks == -1) break;
    for (int j = indeks; j < min(3, indeks + it->second); j++) {
      niz[j] = it->first;
    }
  }
  map<int, int> brojPojavljivanjaUNizu;
  for (int i = 0; i < 3; i++) {
    brojPojavljivanjaUNizu[niz[i]]++;
  }
  long long resenje = 1LL;
  for (map<int, int>::iterator it = brojPojavljivanjaUNizu.begin();
       it != brojPojavljivanjaUNizu.end(); ++it) {
    resenje *= nCr(m[it->first], it->second);
  }
  cout << resenje << '\n';
  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;
int pronadjiPrviIndeks(int niz[3]) {
  for (int i = 0; i < 3; i++) {
    if (niz[i] == -1) return i;
  }
  return -1;
}
long long nCr(long long n, long long k) {
  long long proizvod = 1LL;
  for (int i = 0; i < k; i++) {
    proizvod *= n;
    --n;
  }
  for (int i = 1; i <= k; i++) {
    proizvod /= (long long)i;
  }
  return proizvod;
}
int main() {
  int n;
  cin >> n;
  map<int, int> m;
  for (int i = 0; i < n; i++) {
    int broj;
    cin >> broj;
    ++m[broj];
  }
  int niz[3] = {-1, -1, -1};
  for (map<int, int>::iterator it = m.begin(); it != m.end(); ++it) {
    int indeks = pronadjiPrviIndeks(niz);
    if (indeks == -1) break;
    for (int j = indeks; j < min(3, indeks + it->second); j++) {
      niz[j] = it->first;
    }
  }
  map<int, int> brojPojavljivanjaUNizu;
  for (int i = 0; i < 3; i++) {
    brojPojavljivanjaUNizu[niz[i]]++;
  }
  long long resenje = 1LL;
  for (map<int, int>::iterator it = brojPojavljivanjaUNizu.begin();
       it != brojPojavljivanjaUNizu.end(); ++it) {
    resenje *= nCr(m[it->first], it->second);
  }
  cout << resenje << '\n';
  return 0;
}

```