#include <bits/stdc++.h>
using namespace ::std;
template <class T>
istream& operator>>(istream& I, vector<T>& v) {
  for (T& e : v) I >> e;
  return I;
}
template <class T>
ostream& operator<<(ostream& O, const vector<T>& v) {
  for (const T& e : v) O << e << ' ';
  cout << '\n';
  return O;
}
long long ncr(long long n, long long r) {
  if (r == 1) return n;
  if (r == 2) return n * (n - 1) / 2;
  if (r == 3) return n * (n - 1) * (n - 2) / 6;
}
void _main() {
  long long n;
  cin >> n;
  vector<long long> vec(n);
  cin >> vec;
  sort((vec).begin(), (vec).end());
  map<long long, long long> freq, here;
  for (long long i = 0; i < n; i++) ++freq[vec[i]];
  for (long long i = 0; i < 3; i++) ++here[vec[i]];
  long long ans = 1;
  for (auto it : here) ans *= ncr(freq[it.first], here[it.first]);
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long _t = 1;
  while (_t--) _main();
  return 0;
}

### Prompt

Your task is to create 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;
template <class T>
istream& operator>>(istream& I, vector<T>& v) {
  for (T& e : v) I >> e;
  return I;
}
template <class T>
ostream& operator<<(ostream& O, const vector<T>& v) {
  for (const T& e : v) O << e << ' ';
  cout << '\n';
  return O;
}
long long ncr(long long n, long long r) {
  if (r == 1) return n;
  if (r == 2) return n * (n - 1) / 2;
  if (r == 3) return n * (n - 1) * (n - 2) / 6;
}
void _main() {
  long long n;
  cin >> n;
  vector<long long> vec(n);
  cin >> vec;
  sort((vec).begin(), (vec).end());
  map<long long, long long> freq, here;
  for (long long i = 0; i < n; i++) ++freq[vec[i]];
  for (long long i = 0; i < 3; i++) ++here[vec[i]];
  long long ans = 1;
  for (auto it : here) ans *= ncr(freq[it.first], here[it.first]);
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long _t = 1;
  while (_t--) _main();
  return 0;
}

```