#include <bits/stdc++.h>
using namespace std;
template <class T>
int size(const T &x) {
  return x.size();
}
const int INF = ~(1 << 31);
typedef long int li;
typedef long long int lli;
const double EPS = 1e-9;
const double pi = acos(-1);
template <class T>
T smod(T a, T b) {
  return (a % b + b) % b;
}
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
int main() {
  lli n;
  cin >> n;
  vector<int> ar(n);
  for (__typeof(0) i = (0); i < (100000); ++i) {
    cin >> ar[i];
  }
  sort(ar.begin(), ar.end());
  multiset<lli> st(ar.begin(), ar.end());
  cout << endl;
  lli ans;
  lli n1 = st.count(ar[0]);
  lli i = 1;
  while (ar[i] == ar[0]) i++;
  lli n2 = st.count(ar[i]);
  lli j = i + 1;
  while (ar[j] == ar[i]) j++;
  lli n3 = st.count(ar[j]);
  if (n1 == 1 && n2 == 1) {
    ans = n3;
  } else if (n1 == 1) {
    ans = (n2 * (n2 - 1)) / 2;
  } else if (n1 == 2) {
    ans = n2;
  } else if (n1 >= 3) {
    ans = (n1 * (n1 - 1) * (n1 - 2)) / 6;
  }
  cout << ans;
  return 0;
}

### Prompt

In CPP, your task is to solve 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>
int size(const T &x) {
  return x.size();
}
const int INF = ~(1 << 31);
typedef long int li;
typedef long long int lli;
const double EPS = 1e-9;
const double pi = acos(-1);
template <class T>
T smod(T a, T b) {
  return (a % b + b) % b;
}
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
int main() {
  lli n;
  cin >> n;
  vector<int> ar(n);
  for (__typeof(0) i = (0); i < (100000); ++i) {
    cin >> ar[i];
  }
  sort(ar.begin(), ar.end());
  multiset<lli> st(ar.begin(), ar.end());
  cout << endl;
  lli ans;
  lli n1 = st.count(ar[0]);
  lli i = 1;
  while (ar[i] == ar[0]) i++;
  lli n2 = st.count(ar[i]);
  lli j = i + 1;
  while (ar[j] == ar[i]) j++;
  lli n3 = st.count(ar[j]);
  if (n1 == 1 && n2 == 1) {
    ans = n3;
  } else if (n1 == 1) {
    ans = (n2 * (n2 - 1)) / 2;
  } else if (n1 == 2) {
    ans = n2;
  } else if (n1 >= 3) {
    ans = (n1 * (n1 - 1) * (n1 - 2)) / 6;
  }
  cout << ans;
  return 0;
}

```