#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimization("unroll-loops")
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const long long N = 1e5 + 5;
long long ans, n, a[N];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  map<long long, long long> mp;
  for (long long i = 1; i <= n; i++) mp[a[i]] += 1;
  vector<pair<long long, long long> > v;
  for (pair<long long, long long> ele : mp) v.push_back(ele);
  sort(v.begin(), v.end());
  long long c1 = v[0].second;
  if (c1 >= 3) return cout << (c1 * (c1 - 1) * (c1 - 2)) / 6, 0;
  if (c1 == 2) return cout << v[1].second, 0;
  long long c2 = v[1].second;
  if (c2 >= 2) return cout << (c2 * (c2 - 1)) / 2, 0;
  long long c3 = v[2].second;
  cout << c3;
}

### Prompt

Create a solution in Cpp for 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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimization("unroll-loops")
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const long long N = 1e5 + 5;
long long ans, n, a[N];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  map<long long, long long> mp;
  for (long long i = 1; i <= n; i++) mp[a[i]] += 1;
  vector<pair<long long, long long> > v;
  for (pair<long long, long long> ele : mp) v.push_back(ele);
  sort(v.begin(), v.end());
  long long c1 = v[0].second;
  if (c1 >= 3) return cout << (c1 * (c1 - 1) * (c1 - 2)) / 6, 0;
  if (c1 == 2) return cout << v[1].second, 0;
  long long c2 = v[1].second;
  if (c2 >= 2) return cout << (c2 * (c2 - 1)) / 2, 0;
  long long c3 = v[2].second;
  cout << c3;
}

```