#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const long long INF = 1e18;
const long double PI = 4 * atan((long double)1);
long long N;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> N;
  vector<long long> A(N);
  for (int i = 0; i < (N); i++) cin >> A[i];
  vector<long long> B = A;
  sort(B.begin(), B.end());
  long long a = B[0], b = B[1], c = B[2];
  long long ac = 0, bc = 0, cc = 0;
  for (int i = 0; i < N && (B[i] == a || B[i] == b || B[i] == c); i++) {
    if (B[i] == a) ac++;
    if (B[i] == b) bc++;
    if (B[i] == c) cc++;
  }
  if (a == b && b == c) {
    long long ans = (ac - 2);
    ans = ans * (ac - 1);
    ans = ans * ac;
    ans = ans / 6;
    cout << ans << endl;
  } else if (a == b && b != c) {
    long long ans = ac;
    ans = ans * (ac - 1);
    ans = ans / 2;
    ans = ans * cc;
    cout << ans << endl;
  } else if (b == c && a != b) {
    long long ans = bc;
    ans = ans * (bc - 1);
    ans = ans / 2;
    ans = ans * ac;
    cout << ans << endl;
  } else {
    assert(a != b && b != c);
    long long ans = ac;
    ans = ans * bc;
    ans = ans * cc;
    cout << ans << endl;
  }
}

### 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;
const int MOD = 1000000007;
const long long INF = 1e18;
const long double PI = 4 * atan((long double)1);
long long N;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> N;
  vector<long long> A(N);
  for (int i = 0; i < (N); i++) cin >> A[i];
  vector<long long> B = A;
  sort(B.begin(), B.end());
  long long a = B[0], b = B[1], c = B[2];
  long long ac = 0, bc = 0, cc = 0;
  for (int i = 0; i < N && (B[i] == a || B[i] == b || B[i] == c); i++) {
    if (B[i] == a) ac++;
    if (B[i] == b) bc++;
    if (B[i] == c) cc++;
  }
  if (a == b && b == c) {
    long long ans = (ac - 2);
    ans = ans * (ac - 1);
    ans = ans * ac;
    ans = ans / 6;
    cout << ans << endl;
  } else if (a == b && b != c) {
    long long ans = ac;
    ans = ans * (ac - 1);
    ans = ans / 2;
    ans = ans * cc;
    cout << ans << endl;
  } else if (b == c && a != b) {
    long long ans = bc;
    ans = ans * (bc - 1);
    ans = ans / 2;
    ans = ans * ac;
    cout << ans << endl;
  } else {
    assert(a != b && b != c);
    long long ans = ac;
    ans = ans * bc;
    ans = ans * cc;
    cout << ans << endl;
  }
}

```