#include <bits/stdc++.h>
using namespace std;
template <class T>
void splitstr(const string &s, vector<T> &out) {
  istringstream in(s);
  out.clear();
  copy(istream_iterator<T>(in), istream_iterator<T>(), back_inserter(out));
}
static void redirect(int argc, const char **argv) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  if (argc > 1) {
    static filebuf f;
    f.open(argv[1], ios::in);
    cin.rdbuf(&f);
    if (!cin) {
      cerr << "Failed to open '" << argv[1] << "'" << endl;
      exit(1);
    }
  }
  if (argc > 2) {
    static filebuf f;
    f.open(argv[2], ios::out | ios::trunc);
    cout.rdbuf(&f);
    if (!cout) {
      cerr << "Failed to open '" << argv[2] << "'" << endl;
    }
  }
  cin.exceptions(ios::failbit);
}
static void solve() {
  int N;
  cin >> N;
  vector<long long> a(N);
  for (int i = 0; i < N; i++) cin >> a[i];
  sort(begin(a), end(a));
  long long freq[3];
  for (int i = 0; i < 3; i++) freq[i] = count(begin(a), end(a), a[i]);
  long long ans;
  if (a[0] == a[2])
    ans = freq[0] * (freq[0] - 1) * (freq[0] - 2) / 6;
  else if (a[1] == a[2])
    ans = freq[1] * (freq[1] - 1) / 2;
  else
    ans = freq[2];
  cout << ans << '\n';
}
int main(int argc, const char **argv) {
  redirect(argc, argv);
  solve();
  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;
template <class T>
void splitstr(const string &s, vector<T> &out) {
  istringstream in(s);
  out.clear();
  copy(istream_iterator<T>(in), istream_iterator<T>(), back_inserter(out));
}
static void redirect(int argc, const char **argv) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  if (argc > 1) {
    static filebuf f;
    f.open(argv[1], ios::in);
    cin.rdbuf(&f);
    if (!cin) {
      cerr << "Failed to open '" << argv[1] << "'" << endl;
      exit(1);
    }
  }
  if (argc > 2) {
    static filebuf f;
    f.open(argv[2], ios::out | ios::trunc);
    cout.rdbuf(&f);
    if (!cout) {
      cerr << "Failed to open '" << argv[2] << "'" << endl;
    }
  }
  cin.exceptions(ios::failbit);
}
static void solve() {
  int N;
  cin >> N;
  vector<long long> a(N);
  for (int i = 0; i < N; i++) cin >> a[i];
  sort(begin(a), end(a));
  long long freq[3];
  for (int i = 0; i < 3; i++) freq[i] = count(begin(a), end(a), a[i]);
  long long ans;
  if (a[0] == a[2])
    ans = freq[0] * (freq[0] - 1) * (freq[0] - 2) / 6;
  else if (a[1] == a[2])
    ans = freq[1] * (freq[1] - 1) / 2;
  else
    ans = freq[2];
  cout << ans << '\n';
}
int main(int argc, const char **argv) {
  redirect(argc, argv);
  solve();
  return 0;
}

```