#include <bits/stdc++.h>
using namespace std;
long long int tree[4 * 1000001];
void update(long long int i, long long int l, long long int r,
            long long int index) {
  if (l == r) {
    tree[i] = 1;
    return;
  }
  long long int mid = (l + r) / 2;
  if (mid >= index)
    update(2 * i, l, mid, index);
  else
    update(2 * i + 1, mid + 1, r, index);
  tree[i] = tree[2 * i] + tree[2 * i + 1];
}
long long int query(long long int i, long long int l, long long int r,
                    long long int ql, long long int qr) {
  if (ql <= l && qr >= r) return tree[i];
  if (ql > r || qr < l) return 0;
  long long int mid = (l + r) / 2;
  return query(2 * i, l, mid, ql, qr) + query(2 * i + 1, mid + 1, r, ql, qr);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  vector<pair<long long int, long long int> > A(n);
  for (int i = 0; i < n; i++) {
    cin >> A[i].first;
    A[i].second = i;
  }
  sort(A.begin(), A.end());
  reverse(A.begin(), A.end());
  vector<long long int> left(n, 0), right(n, 0);
  for (int i = 0; i < n; i++) {
    long long int index = A[i].second;
    left[index] = query(1, 0, n - 1, 0, max(0LL, index - 1));
    right[index] = n - 1 + left[index] - index - i;
    update(1, 0, n - 1, index);
  }
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    ans += (left[i] * right[i]);
  }
  cout << ans;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
long long int tree[4 * 1000001];
void update(long long int i, long long int l, long long int r,
            long long int index) {
  if (l == r) {
    tree[i] = 1;
    return;
  }
  long long int mid = (l + r) / 2;
  if (mid >= index)
    update(2 * i, l, mid, index);
  else
    update(2 * i + 1, mid + 1, r, index);
  tree[i] = tree[2 * i] + tree[2 * i + 1];
}
long long int query(long long int i, long long int l, long long int r,
                    long long int ql, long long int qr) {
  if (ql <= l && qr >= r) return tree[i];
  if (ql > r || qr < l) return 0;
  long long int mid = (l + r) / 2;
  return query(2 * i, l, mid, ql, qr) + query(2 * i + 1, mid + 1, r, ql, qr);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  vector<pair<long long int, long long int> > A(n);
  for (int i = 0; i < n; i++) {
    cin >> A[i].first;
    A[i].second = i;
  }
  sort(A.begin(), A.end());
  reverse(A.begin(), A.end());
  vector<long long int> left(n, 0), right(n, 0);
  for (int i = 0; i < n; i++) {
    long long int index = A[i].second;
    left[index] = query(1, 0, n - 1, 0, max(0LL, index - 1));
    right[index] = n - 1 + left[index] - index - i;
    update(1, 0, n - 1, index);
  }
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    ans += (left[i] * right[i]);
  }
  cout << ans;
  return 0;
}

```