#include <bits/stdc++.h>
using namespace std;
const int mx = 100010;
int a[mx], n;
class ui_t {
 public:
  unsigned long long m[4];
  ui_t(unsigned long long x) {
    m[0] = x % 0x100000000;
    m[1] = x / 0x100000000;
    m[2] = 0;
    m[3] = 0;
  }
  ui_t& operator*=(unsigned long long x) {
    unsigned long long keep = 0;
    for (int i = 0; i < 4; i++) {
      m[i] = keep + m[i] * x;
      keep = m[i] / 0x100000000;
      m[i] %= 0x100000000;
    }
    return *this;
  }
  bool operator<(const ui_t& o) const {
    for (int i = 3; i >= 0; i--) {
      if (m[i] < o.m[i]) return true;
      if (m[i] > o.m[i]) return false;
    }
    return false;
  }
  bool operator==(const ui_t& o) const {
    for (int i = 0; i < 4; i++) {
      if (m[i] != o.m[i]) return false;
    }
    return true;
  }
};
struct TFeatures {
  int leftmin, rightmin;
  int leftmin_cnt, rightmin_cnt;
} features[mx];
int main(int argc, char* argv[]) {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int mmin, nmin;
  int leftmin = a[1];
  int leftmin_cnt = 1;
  for (int i = 2; i < n; i++) {
    features[i].leftmin = leftmin;
    features[i].leftmin_cnt = leftmin_cnt;
    if (a[i] < leftmin) {
      leftmin = a[i];
      leftmin_cnt = 1;
    } else if (a[i] == leftmin) {
      leftmin_cnt++;
    }
  }
  int rightmin = a[n];
  int rightmin_cnt = 1;
  for (int i = n - 1; i > 1; i--) {
    features[i].rightmin = rightmin;
    features[i].rightmin_cnt = rightmin_cnt;
    if (a[i] < rightmin) {
      rightmin = a[i];
      rightmin_cnt = 1;
    } else if (a[i] == rightmin) {
      rightmin_cnt++;
    }
  }
  long long minProd = -1;
  long long r = 0;
  ui_t mi(0);
  for (int i = 2; i < n; i++) {
    ui_t m(features[i].leftmin);
    m *= a[i];
    m *= features[i].rightmin;
    if (i == 2 || m < mi)
      mi = m, r = 1ll * features[i].leftmin_cnt * features[i].rightmin_cnt;
    else if (m == mi)
      r += 1ll * features[i].leftmin_cnt * features[i].rightmin_cnt;
  }
  printf("%lld\n", r);
  return 0;
}

### Prompt

Your challenge is to write 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 mx = 100010;
int a[mx], n;
class ui_t {
 public:
  unsigned long long m[4];
  ui_t(unsigned long long x) {
    m[0] = x % 0x100000000;
    m[1] = x / 0x100000000;
    m[2] = 0;
    m[3] = 0;
  }
  ui_t& operator*=(unsigned long long x) {
    unsigned long long keep = 0;
    for (int i = 0; i < 4; i++) {
      m[i] = keep + m[i] * x;
      keep = m[i] / 0x100000000;
      m[i] %= 0x100000000;
    }
    return *this;
  }
  bool operator<(const ui_t& o) const {
    for (int i = 3; i >= 0; i--) {
      if (m[i] < o.m[i]) return true;
      if (m[i] > o.m[i]) return false;
    }
    return false;
  }
  bool operator==(const ui_t& o) const {
    for (int i = 0; i < 4; i++) {
      if (m[i] != o.m[i]) return false;
    }
    return true;
  }
};
struct TFeatures {
  int leftmin, rightmin;
  int leftmin_cnt, rightmin_cnt;
} features[mx];
int main(int argc, char* argv[]) {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int mmin, nmin;
  int leftmin = a[1];
  int leftmin_cnt = 1;
  for (int i = 2; i < n; i++) {
    features[i].leftmin = leftmin;
    features[i].leftmin_cnt = leftmin_cnt;
    if (a[i] < leftmin) {
      leftmin = a[i];
      leftmin_cnt = 1;
    } else if (a[i] == leftmin) {
      leftmin_cnt++;
    }
  }
  int rightmin = a[n];
  int rightmin_cnt = 1;
  for (int i = n - 1; i > 1; i--) {
    features[i].rightmin = rightmin;
    features[i].rightmin_cnt = rightmin_cnt;
    if (a[i] < rightmin) {
      rightmin = a[i];
      rightmin_cnt = 1;
    } else if (a[i] == rightmin) {
      rightmin_cnt++;
    }
  }
  long long minProd = -1;
  long long r = 0;
  ui_t mi(0);
  for (int i = 2; i < n; i++) {
    ui_t m(features[i].leftmin);
    m *= a[i];
    m *= features[i].rightmin;
    if (i == 2 || m < mi)
      mi = m, r = 1ll * features[i].leftmin_cnt * features[i].rightmin_cnt;
    else if (m == mi)
      r += 1ll * features[i].leftmin_cnt * features[i].rightmin_cnt;
  }
  printf("%lld\n", r);
  return 0;
}

```