#include <bits/stdc++.h>
using namespace std;
int arr[1000001], temp[1000001];
int BITree[1000002];
int getFrontSum(int index) {
  int sum = 0;
  while (index > 0) {
    sum += BITree[index];
    index -= (index & (-index));
  }
  return sum;
}
int getBackSum(int index, int n) {
  int sum = 0;
  while (index <= n) {
    sum += BITree[index];
    index += (index & (-index));
  }
  return sum;
}
void Update(int index, int vl, int n) {
  while (index <= n) {
    BITree[index] += vl;
    index += (index & (-index));
  }
}
int main() {
  int n;
  cin >> n;
  int i;
  memset(BITree, 0, sizeof(BITree));
  long long int smaller[n], greater[n];
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    temp[i] = arr[i];
  }
  sort(temp, temp + n);
  for (i = 0; i < n; i++)
    arr[i] = (lower_bound(temp, temp + n, arr[i]) - temp + 1);
  long long int cnt = 0;
  for (i = n - 1; i >= 0; i--) {
    smaller[i] = (getFrontSum(arr[i] - 1));
    Update(arr[i], 1, n);
  }
  memset(BITree, 0, sizeof(BITree));
  for (i = 0; i < n; i++) {
    greater[i] = i - getFrontSum(arr[i]);
    Update(arr[i], 1, n);
  }
  for (i = 0; i < n; i++) cnt += (smaller[i] * greater[i]);
  cout << cnt << "\n";
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

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;
int arr[1000001], temp[1000001];
int BITree[1000002];
int getFrontSum(int index) {
  int sum = 0;
  while (index > 0) {
    sum += BITree[index];
    index -= (index & (-index));
  }
  return sum;
}
int getBackSum(int index, int n) {
  int sum = 0;
  while (index <= n) {
    sum += BITree[index];
    index += (index & (-index));
  }
  return sum;
}
void Update(int index, int vl, int n) {
  while (index <= n) {
    BITree[index] += vl;
    index += (index & (-index));
  }
}
int main() {
  int n;
  cin >> n;
  int i;
  memset(BITree, 0, sizeof(BITree));
  long long int smaller[n], greater[n];
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    temp[i] = arr[i];
  }
  sort(temp, temp + n);
  for (i = 0; i < n; i++)
    arr[i] = (lower_bound(temp, temp + n, arr[i]) - temp + 1);
  long long int cnt = 0;
  for (i = n - 1; i >= 0; i--) {
    smaller[i] = (getFrontSum(arr[i] - 1));
    Update(arr[i], 1, n);
  }
  memset(BITree, 0, sizeof(BITree));
  for (i = 0; i < n; i++) {
    greater[i] = i - getFrontSum(arr[i]);
    Update(arr[i], 1, n);
  }
  for (i = 0; i < n; i++) cnt += (smaller[i] * greater[i]);
  cout << cnt << "\n";
  return 0;
}

```