#include <bits/stdc++.h>
const int kMaxN = 1000000;
int a[kMaxN], x[kMaxN];
long long f[kMaxN + 1], occ[kMaxN + 1];
int n;
void ReadIn() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    x[i] = a[i];
  }
  std::sort(x, x + n);
}
void Insert(long long *f, int x, long long t) {
  while (x <= n) {
    f[x] += t;
    x += x & (x ^ (x - 1));
  }
}
long long Find(long long *f, int x) {
  long long result = 0;
  while (x > 0) {
    result += f[x];
    x -= x & (x ^ (x - 1));
  }
  return result;
}
void Solve() {
  long long result = 0;
  for (int i = 0; i < n; ++i) {
    int id = n - (std::lower_bound(x, x + n, a[i]) - x);
    result += Find(f, id - 1);
    Insert(f, id, Find(occ, id - 1));
    Insert(occ, id, 1);
  }
  printf("%I64d\n", result);
}
int main() {
  ReadIn();
  Solve();
  return 0;
}

### 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>
const int kMaxN = 1000000;
int a[kMaxN], x[kMaxN];
long long f[kMaxN + 1], occ[kMaxN + 1];
int n;
void ReadIn() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    x[i] = a[i];
  }
  std::sort(x, x + n);
}
void Insert(long long *f, int x, long long t) {
  while (x <= n) {
    f[x] += t;
    x += x & (x ^ (x - 1));
  }
}
long long Find(long long *f, int x) {
  long long result = 0;
  while (x > 0) {
    result += f[x];
    x -= x & (x ^ (x - 1));
  }
  return result;
}
void Solve() {
  long long result = 0;
  for (int i = 0; i < n; ++i) {
    int id = n - (std::lower_bound(x, x + n, a[i]) - x);
    result += Find(f, id - 1);
    Insert(f, id, Find(occ, id - 1));
    Insert(occ, id, 1);
  }
  printf("%I64d\n", result);
}
int main() {
  ReadIn();
  Solve();
  return 0;
}

```