#include <bits/stdc++.h>
using namespace std;
long long int bit[1000009];
long long int query(long long int i) {
  int ans = 0;
  for (; i > 0; i -= i & -i) ans += bit[i];
  return ans;
}
void update(long long int i, long long int x) {
  for (; i < 1000009; i += i & -i) bit[i] += x;
}
int main() {
  long long int n;
  cin >> n;
  long long int a[n], b[n];
  for (int i = 0; i < n; i++) {
    long long int x;
    cin >> x;
    a[i] = b[i] = x;
  }
  sort(b, b + n);
  unordered_map<long long int, long long int> m;
  long long int k = 1;
  for (int i = 0; i < n; i++) m[b[i]] = k++;
  for (int i = 0; i < n; i++) a[i] = m[a[i]];
  long long int ans = 0, aux[n + 3], oporra[n + 3];
  for (int i = 0; i < n; i++) {
    aux[i] = query(n) - query(a[i]);
    oporra[i] = query(a[i] - 1);
    update(a[i], 1);
  }
  for (int i = 0; i < n; i++) {
    long long int d = (query(a[i] - 1) - oporra[i]);
    ans += d * aux[i];
  }
  cout << ans << endl;
}

### Prompt

Please create a solution in Cpp 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 int bit[1000009];
long long int query(long long int i) {
  int ans = 0;
  for (; i > 0; i -= i & -i) ans += bit[i];
  return ans;
}
void update(long long int i, long long int x) {
  for (; i < 1000009; i += i & -i) bit[i] += x;
}
int main() {
  long long int n;
  cin >> n;
  long long int a[n], b[n];
  for (int i = 0; i < n; i++) {
    long long int x;
    cin >> x;
    a[i] = b[i] = x;
  }
  sort(b, b + n);
  unordered_map<long long int, long long int> m;
  long long int k = 1;
  for (int i = 0; i < n; i++) m[b[i]] = k++;
  for (int i = 0; i < n; i++) a[i] = m[a[i]];
  long long int ans = 0, aux[n + 3], oporra[n + 3];
  for (int i = 0; i < n; i++) {
    aux[i] = query(n) - query(a[i]);
    oporra[i] = query(a[i] - 1);
    update(a[i], 1);
  }
  for (int i = 0; i < n; i++) {
    long long int d = (query(a[i] - 1) - oporra[i]);
    ans += d * aux[i];
  }
  cout << ans << endl;
}

```