#include <bits/stdc++.h>
long long int mod = 1000000007;
using namespace std;
pair<int, int> arr[1000000];
pair<int, int> b[1000000];
int cnt[1000000];
int cnt2[1000000];
int ct = 0;
void merge(int l, int mid, int r) {
  pair<int, int> a[mid - l + 1], b[r - mid];
  int tmp[mid - l + 1];
  for (int i = 0; i < mid - l + 1; i++) {
    a[i] = arr[l + i];
    tmp[i] = 0;
  }
  for (int i = 0; i < r - mid; i++) {
    b[i] = arr[mid + 1 + i];
  }
  int i = 0, j = 0, k = l;
  while (i < mid - l + 1 && j < r - mid) {
    if (a[i].first < b[j].first) {
      arr[k++] = a[i++];
      cnt2[a[i - 1].second] += j;
    } else if (a[i].first == b[j].first) {
      arr[k++] = a[i++];
    } else {
      arr[k++] = b[j++];
      ct = ct + mid - i + 1;
      ct = ct - l;
      cnt[b[j - 1].second] += mid - i + 1 - l;
    }
  }
  while (i < mid - l + 1) {
    arr[k++] = a[i++];
    cnt2[a[i - 1].second] += r - mid;
  }
  while (j < r - mid) {
    arr[k++] = b[j++];
  }
}
void mergesort(int l, int r) {
  if (l < r) {
    int mid = (l + r) / 2;
    mergesort(l, mid);
    mergesort(mid + 1, r);
    merge(l, mid, r);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i].first;
    arr[i].second = i;
  }
  mergesort(0, n - 1);
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    ans += (long long int)cnt[i] * (long long int)cnt2[i];
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create 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>
long long int mod = 1000000007;
using namespace std;
pair<int, int> arr[1000000];
pair<int, int> b[1000000];
int cnt[1000000];
int cnt2[1000000];
int ct = 0;
void merge(int l, int mid, int r) {
  pair<int, int> a[mid - l + 1], b[r - mid];
  int tmp[mid - l + 1];
  for (int i = 0; i < mid - l + 1; i++) {
    a[i] = arr[l + i];
    tmp[i] = 0;
  }
  for (int i = 0; i < r - mid; i++) {
    b[i] = arr[mid + 1 + i];
  }
  int i = 0, j = 0, k = l;
  while (i < mid - l + 1 && j < r - mid) {
    if (a[i].first < b[j].first) {
      arr[k++] = a[i++];
      cnt2[a[i - 1].second] += j;
    } else if (a[i].first == b[j].first) {
      arr[k++] = a[i++];
    } else {
      arr[k++] = b[j++];
      ct = ct + mid - i + 1;
      ct = ct - l;
      cnt[b[j - 1].second] += mid - i + 1 - l;
    }
  }
  while (i < mid - l + 1) {
    arr[k++] = a[i++];
    cnt2[a[i - 1].second] += r - mid;
  }
  while (j < r - mid) {
    arr[k++] = b[j++];
  }
}
void mergesort(int l, int r) {
  if (l < r) {
    int mid = (l + r) / 2;
    mergesort(l, mid);
    mergesort(mid + 1, r);
    merge(l, mid, r);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i].first;
    arr[i].second = i;
  }
  mergesort(0, n - 1);
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    ans += (long long int)cnt[i] * (long long int)cnt2[i];
  }
  cout << ans << endl;
  return 0;
}

```