#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long double PI = acos(-1);
const long long MOD = 1000000007;
const long long FMOD = 998244353;
const long double eps = 1e-9;
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());
signed main() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  map<long long, long long> mp1;
  for (long long i = 0; i < n; i++) {
    mp1[a[i]]++;
  }
  long long ans = 0, cnt = 0;
  set<long long> st;
  for (auto x : mp1) {
    if (cnt >= 3) break;
    st.insert(x.first);
    cnt += x.second;
  }
  if (st.size() == 1) {
    long long si = mp1[*st.begin()];
    cout << si * (si - 1) * (si - 2) / 6;
  } else if (st.size() == 2) {
    long long si1 = mp1[*st.begin()];
    long long si2 = mp1[*(--st.end())];
    if (si1 >= 2)
      cout << si1 * (si1 - 1) / 2 * si2;
    else if (si2 >= 2)
      cout << si1 * si2 * (si2 - 1) / 2;
  } else {
    cout << mp1[*(--st.end())];
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 GCC optimize("O3")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long double PI = acos(-1);
const long long MOD = 1000000007;
const long long FMOD = 998244353;
const long double eps = 1e-9;
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());
signed main() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  map<long long, long long> mp1;
  for (long long i = 0; i < n; i++) {
    mp1[a[i]]++;
  }
  long long ans = 0, cnt = 0;
  set<long long> st;
  for (auto x : mp1) {
    if (cnt >= 3) break;
    st.insert(x.first);
    cnt += x.second;
  }
  if (st.size() == 1) {
    long long si = mp1[*st.begin()];
    cout << si * (si - 1) * (si - 2) / 6;
  } else if (st.size() == 2) {
    long long si1 = mp1[*st.begin()];
    long long si2 = mp1[*(--st.end())];
    if (si1 >= 2)
      cout << si1 * (si1 - 1) / 2 * si2;
    else if (si2 >= 2)
      cout << si1 * si2 * (si2 - 1) / 2;
  } else {
    cout << mp1[*(--st.end())];
  }
}

```