#include <bits/stdc++.h>
using namespace std;
vector<pair<long long, long long>> calcMin(const vector<long long> &v,
                                           bool l2r) {
  pair<long long, long long> min = {2e9, 0};
  vector<pair<long long, long long>> res;
  auto f = [&min](long long x) -> pair<long long, long long> {
    if (x < min.first) {
      min.first = x;
      min.second = 1;
    } else if (x == min.first) {
      min.second++;
    }
    return min;
  };
  if (!l2r)
    std::transform(v.begin(), v.end(), std::back_inserter(res), f);
  else {
    std::transform(v.rbegin(), v.rend(), std::back_inserter(res), f);
    std::reverse(res.begin(), res.end());
  }
  return res;
}
multiset<long long> findMinProd(vector<long long> v) {
  sort(v.begin(), v.end());
  multiset<long long> s = {v[0], v[1], v[2]};
  return s;
}
int main() {
  int n;
  cin >> n;
  vector<long long> v(n, 0);
  for (auto &e : v) {
    cin >> e;
  }
  vector<pair<long long, long long>> l = calcMin(v, 0), r = calcMin(v, 1);
  auto prodMin = findMinProd(v);
  long long res = 0;
  for (int j = 1; j < n - 1; j++) {
    multiset<long long> s{v[j], l[j - 1].first, r[j + 1].first};
    if (s == prodMin) {
      res += l[j - 1].second * r[j + 1].second;
    }
  }
  cout << res;
  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;
vector<pair<long long, long long>> calcMin(const vector<long long> &v,
                                           bool l2r) {
  pair<long long, long long> min = {2e9, 0};
  vector<pair<long long, long long>> res;
  auto f = [&min](long long x) -> pair<long long, long long> {
    if (x < min.first) {
      min.first = x;
      min.second = 1;
    } else if (x == min.first) {
      min.second++;
    }
    return min;
  };
  if (!l2r)
    std::transform(v.begin(), v.end(), std::back_inserter(res), f);
  else {
    std::transform(v.rbegin(), v.rend(), std::back_inserter(res), f);
    std::reverse(res.begin(), res.end());
  }
  return res;
}
multiset<long long> findMinProd(vector<long long> v) {
  sort(v.begin(), v.end());
  multiset<long long> s = {v[0], v[1], v[2]};
  return s;
}
int main() {
  int n;
  cin >> n;
  vector<long long> v(n, 0);
  for (auto &e : v) {
    cin >> e;
  }
  vector<pair<long long, long long>> l = calcMin(v, 0), r = calcMin(v, 1);
  auto prodMin = findMinProd(v);
  long long res = 0;
  for (int j = 1; j < n - 1; j++) {
    multiset<long long> s{v[j], l[j - 1].first, r[j + 1].first};
    if (s == prodMin) {
      res += l[j - 1].second * r[j + 1].second;
    }
  }
  cout << res;
  return 0;
}

```