#include <bits/stdc++.h>
using namespace std;
void mergesort(vector<int> &, int, int);
vector<long int> Num;
vector<int> A, B;
int main() {
  int n;
  cin >> n;
  vector<int> index;
  A.resize(n + 1, 0);
  B.resize(n + 1, 0);
  Num.push_back(0);
  for (int i = 1; i <= n; i++) {
    int temp;
    cin >> temp;
    Num.push_back(temp);
  }
  if (n <= 2) {
    cout << "0" << endl;
    return 0;
  }
  index.push_back(0);
  for (int i = 1; i <= n; i++) {
    index.push_back(i);
  }
  mergesort(index, 1, n);
  long long int answer = 0;
  for (int i = 2; i <= n - 1; i++) {
    answer += (long long)A[i] * B[i];
  }
  cout << answer << endl;
}
void mergesort(vector<int> &index, int left, int right) {
  if (right - left < 1) return;
  int mid = (left + right) / 2;
  mergesort(index, left, mid);
  mergesort(index, mid + 1, right);
  vector<int> a, b;
  for (int i = left; i <= mid; i++) a.push_back(index[i]);
  for (int i = mid + 1; i <= right; i++) b.push_back(index[i]);
  int i = 0, j = 0, now;
  for (now = left; now <= right; now++) {
    if (i >= a.size()) {
      index[now] = b[j];
      j++;
    } else if (j >= b.size()) {
      A[a[i]] += b.size();
      index[now] = a[i];
      i++;
    } else if (Num[a[i]] <= Num[b[j]]) {
      A[a[i]] += j;
      index[now] = a[i];
      i++;
    } else {
      B[b[j]] += (a.size() - i);
      index[now] = b[j];
      j++;
    }
  }
}

### Prompt

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

The Romans have attacked again. This time they are much more than the Persians but Shapur is ready to defeat them. He says: "A lion is never afraid of a hundred sheep". 

Nevertheless Shapur has to find weaknesses in the Roman army to defeat them. So he gives the army a weakness number.

In Shapur's opinion the weakness of an army is equal to the number of triplets i, j, k such that i < j < k and ai > aj > ak where ax is the power of man standing at position x. The Roman army has one special trait — powers of all the people in it are distinct.

Help Shapur find out how weak the Romans are.

Input

The first line of input contains a single number n (3 ≤ n ≤ 106) — the number of men in Roman army. Next line contains n different positive integers ai (1 ≤ i ≤ n, 1 ≤ ai ≤ 109) — powers of men in the Roman army. 

Output

A single integer number, the weakness of the Roman army. 

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Examples

Input

3
3 2 1


Output

1


Input

3
2 3 1


Output

0


Input

4
10 8 3 1


Output

4


Input

4
1 5 4 3


Output

1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void mergesort(vector<int> &, int, int);
vector<long int> Num;
vector<int> A, B;
int main() {
  int n;
  cin >> n;
  vector<int> index;
  A.resize(n + 1, 0);
  B.resize(n + 1, 0);
  Num.push_back(0);
  for (int i = 1; i <= n; i++) {
    int temp;
    cin >> temp;
    Num.push_back(temp);
  }
  if (n <= 2) {
    cout << "0" << endl;
    return 0;
  }
  index.push_back(0);
  for (int i = 1; i <= n; i++) {
    index.push_back(i);
  }
  mergesort(index, 1, n);
  long long int answer = 0;
  for (int i = 2; i <= n - 1; i++) {
    answer += (long long)A[i] * B[i];
  }
  cout << answer << endl;
}
void mergesort(vector<int> &index, int left, int right) {
  if (right - left < 1) return;
  int mid = (left + right) / 2;
  mergesort(index, left, mid);
  mergesort(index, mid + 1, right);
  vector<int> a, b;
  for (int i = left; i <= mid; i++) a.push_back(index[i]);
  for (int i = mid + 1; i <= right; i++) b.push_back(index[i]);
  int i = 0, j = 0, now;
  for (now = left; now <= right; now++) {
    if (i >= a.size()) {
      index[now] = b[j];
      j++;
    } else if (j >= b.size()) {
      A[a[i]] += b.size();
      index[now] = a[i];
      i++;
    } else if (Num[a[i]] <= Num[b[j]]) {
      A[a[i]] += j;
      index[now] = a[i];
      i++;
    } else {
      B[b[j]] += (a.size() - i);
      index[now] = b[j];
      j++;
    }
  }
}

```