#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
int n;
int a[N];
int m1, m2, m3;
int x1, x2;
int l1[N], r1[N];
int l2[N], r2[N];
int l3[N], r3[N];
long long solve(int m, int l[], int r[]) {
  long long res = 0;
  for (int i = 2; i < n; i++)
    if (a[i] == m) res += (long long)l[i - 1] * r[i + 1];
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  m1 = m2 = m3 = 1e9;
  for (int i = 1; i <= n; i++)
    if (m1 > a[i]) m1 = a[i], x1 = i;
  for (int i = 1; i <= n; i++)
    if (i != x1 && m2 > a[i]) m2 = a[i], x2 = i;
  for (int i = 1; i <= n; i++)
    if (i != x1 && i != x2 && m3 > a[i]) m3 = a[i];
  for (int i = 1; i <= n; i++) {
    l1[i] = l1[i - 1] + (a[i] == m1);
    l2[i] = l2[i - 1] + (a[i] == m2);
    l3[i] = l3[i - 1] + (a[i] == m3);
  }
  for (int i = n; i >= 1; i--) {
    r1[i] = r1[i + 1] + (a[i] == m1);
    r2[i] = r2[i + 1] + (a[i] == m2);
    r3[i] = r3[i + 1] + (a[i] == m3);
  }
  long long ans = 0;
  if (m1 == m2 && m2 == m3)
    ans += solve(m2, l1, r3);
  else if (m1 == m2) {
    ans += solve(m2, l1, r3);
    ans += solve(m1, l3, r2);
    ans += solve(m3, l1, r2);
  } else if (m2 == m3) {
    ans += solve(m2, l1, r3);
    ans += solve(m1, l2, r3);
    ans += solve(m3, l2, r1);
  } else {
    ans += solve(m2, l1, r3);
    ans += solve(m3, l1, r2);
    ans += solve(m1, l2, r3);
    ans += solve(m3, l2, r1);
    ans += solve(m1, l3, r2);
    ans += solve(m2, l3, r1);
  }
  printf("%I64d\n", ans);
}

### 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 int N = 100005;
int n;
int a[N];
int m1, m2, m3;
int x1, x2;
int l1[N], r1[N];
int l2[N], r2[N];
int l3[N], r3[N];
long long solve(int m, int l[], int r[]) {
  long long res = 0;
  for (int i = 2; i < n; i++)
    if (a[i] == m) res += (long long)l[i - 1] * r[i + 1];
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  m1 = m2 = m3 = 1e9;
  for (int i = 1; i <= n; i++)
    if (m1 > a[i]) m1 = a[i], x1 = i;
  for (int i = 1; i <= n; i++)
    if (i != x1 && m2 > a[i]) m2 = a[i], x2 = i;
  for (int i = 1; i <= n; i++)
    if (i != x1 && i != x2 && m3 > a[i]) m3 = a[i];
  for (int i = 1; i <= n; i++) {
    l1[i] = l1[i - 1] + (a[i] == m1);
    l2[i] = l2[i - 1] + (a[i] == m2);
    l3[i] = l3[i - 1] + (a[i] == m3);
  }
  for (int i = n; i >= 1; i--) {
    r1[i] = r1[i + 1] + (a[i] == m1);
    r2[i] = r2[i + 1] + (a[i] == m2);
    r3[i] = r3[i + 1] + (a[i] == m3);
  }
  long long ans = 0;
  if (m1 == m2 && m2 == m3)
    ans += solve(m2, l1, r3);
  else if (m1 == m2) {
    ans += solve(m2, l1, r3);
    ans += solve(m1, l3, r2);
    ans += solve(m3, l1, r2);
  } else if (m2 == m3) {
    ans += solve(m2, l1, r3);
    ans += solve(m1, l2, r3);
    ans += solve(m3, l2, r1);
  } else {
    ans += solve(m2, l1, r3);
    ans += solve(m3, l1, r2);
    ans += solve(m1, l2, r3);
    ans += solve(m3, l2, r1);
    ans += solve(m1, l3, r2);
    ans += solve(m2, l3, r1);
  }
  printf("%I64d\n", ans);
}

```