#include <bits/stdc++.h>
using namespace std;
const unsigned long long M = 1e9 + 10;
const unsigned long long N = 1e9 + 10;
map<unsigned long long, unsigned long long> cnt;
unsigned long long combination(unsigned long long n, unsigned long long m) {
  unsigned long long i, j;
  unsigned long long sum = 1;
  if (n == 0 || m == 0)
    return 1;
  else {
    for (i = m + 1, j = 1; i <= n; i++, j++) {
      sum = sum * i / j;
    }
    return sum;
  }
}
int main() {
  unsigned long long n;
  cin >> n;
  unsigned long long m1 = N, m2 = N, m3 = N;
  for (unsigned long long i = 0; i < n; i++) {
    unsigned long long t;
    scanf("%lld", &t);
    if (t < m1) {
      m3 = m2;
      m2 = m1;
      m1 = t;
    } else if (t < m2) {
      m3 = m2;
      m2 = t;
    } else if (t < m3) {
      m3 = t;
    }
    cnt[t]++;
  }
  if (m1 == m2 && m1 == m3)
    cout << combination(cnt[m1], 3) << endl;
  else if ((m1 == m2 && m1 != m3))
    cout << combination(cnt[m1], 2) * cnt[m3] << endl;
  else if ((m1 == m3 && m1 != m2))
    cout << combination(cnt[m1], 2) * cnt[m2] << endl;
  else if ((m2 == m3 && m1 != m2))
    cout << cnt[m1] * combination(cnt[m2], 2) << endl;
  else
    cout << cnt[m1] * cnt[m2] * cnt[m3] << endl;
}

### Prompt

Please formulate 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;
const unsigned long long M = 1e9 + 10;
const unsigned long long N = 1e9 + 10;
map<unsigned long long, unsigned long long> cnt;
unsigned long long combination(unsigned long long n, unsigned long long m) {
  unsigned long long i, j;
  unsigned long long sum = 1;
  if (n == 0 || m == 0)
    return 1;
  else {
    for (i = m + 1, j = 1; i <= n; i++, j++) {
      sum = sum * i / j;
    }
    return sum;
  }
}
int main() {
  unsigned long long n;
  cin >> n;
  unsigned long long m1 = N, m2 = N, m3 = N;
  for (unsigned long long i = 0; i < n; i++) {
    unsigned long long t;
    scanf("%lld", &t);
    if (t < m1) {
      m3 = m2;
      m2 = m1;
      m1 = t;
    } else if (t < m2) {
      m3 = m2;
      m2 = t;
    } else if (t < m3) {
      m3 = t;
    }
    cnt[t]++;
  }
  if (m1 == m2 && m1 == m3)
    cout << combination(cnt[m1], 3) << endl;
  else if ((m1 == m2 && m1 != m3))
    cout << combination(cnt[m1], 2) * cnt[m3] << endl;
  else if ((m1 == m3 && m1 != m2))
    cout << combination(cnt[m1], 2) * cnt[m2] << endl;
  else if ((m2 == m3 && m1 != m2))
    cout << cnt[m1] * combination(cnt[m2], 2) << endl;
  else
    cout << cnt[m1] * cnt[m2] * cnt[m3] << endl;
}

```