#include <bits/stdc++.h>
int c[1000001], n, a[1000001], orig[1000001];
long long x[1000001], ans;
void update(int ind) {
  for (; ind <= n; ind += ind & -ind) ++c[ind];
}
int query(int ind) {
  int tot = 0;
  for (; ind; ind -= ind & -ind) tot += c[ind];
  return tot;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i), orig[i] = a[i];
  std::sort(orig + 1, orig + n + 1);
  for (int i = 1; i <= n; i++)
    a[i] = std::lower_bound(orig + 1, orig + n + 1, a[i]) - orig;
  for (int i = n; i; i--) {
    x[i] = query(a[i]);
    update(a[i]);
  }
  memset(c, 0, sizeof c);
  for (int i = 1; i <= n; i++) {
    ans += x[i] * query(n - a[i] + 1);
    update(n - a[i] + 1);
  }
  printf("%I64d", ans);
}

### 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>
int c[1000001], n, a[1000001], orig[1000001];
long long x[1000001], ans;
void update(int ind) {
  for (; ind <= n; ind += ind & -ind) ++c[ind];
}
int query(int ind) {
  int tot = 0;
  for (; ind; ind -= ind & -ind) tot += c[ind];
  return tot;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i), orig[i] = a[i];
  std::sort(orig + 1, orig + n + 1);
  for (int i = 1; i <= n; i++)
    a[i] = std::lower_bound(orig + 1, orig + n + 1, a[i]) - orig;
  for (int i = n; i; i--) {
    x[i] = query(a[i]);
    update(a[i]);
  }
  memset(c, 0, sizeof c);
  for (int i = 1; i <= n; i++) {
    ans += x[i] * query(n - a[i] + 1);
    update(n - a[i] + 1);
  }
  printf("%I64d", ans);
}

```