#include <bits/stdc++.h>
using namespace std;
long long tree[2][1000001];
int n;
pair<long long, long long> a[1000001];
int b[1000001];
long long find(int ind, int t) {
  if (ind == 0) return 0;
  return (tree[t][ind] + find(ind - (ind & -ind), t));
}
void insert(int ind, long long num, int t) {
  if (ind > 1000000) return;
  tree[t][ind] += num;
  insert(ind + (ind & -ind), num, t);
}
int main() {
  int size = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    if ((i == 0) || (a[i].first != a[i - 1].first)) {
      size++;
      b[a[i].second] = size;
    }
  }
  long long res = 0;
  long long tmp;
  for (int i = n - 1; i >= 0; i--) {
    insert(b[i], 1, 0);
    res += find(b[i] - 1, 1);
    tmp = find(b[i] - 1, 0);
    insert(b[i], tmp, 1);
  }
  cout << res << endl;
}

### Prompt

Your challenge is to write 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;
long long tree[2][1000001];
int n;
pair<long long, long long> a[1000001];
int b[1000001];
long long find(int ind, int t) {
  if (ind == 0) return 0;
  return (tree[t][ind] + find(ind - (ind & -ind), t));
}
void insert(int ind, long long num, int t) {
  if (ind > 1000000) return;
  tree[t][ind] += num;
  insert(ind + (ind & -ind), num, t);
}
int main() {
  int size = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    if ((i == 0) || (a[i].first != a[i - 1].first)) {
      size++;
      b[a[i].second] = size;
    }
  }
  long long res = 0;
  long long tmp;
  for (int i = n - 1; i >= 0; i--) {
    insert(b[i], 1, 0);
    res += find(b[i] - 1, 1);
    tmp = find(b[i] - 1, 0);
    insert(b[i], tmp, 1);
  }
  cout << res << endl;
}

```