#include <bits/stdc++.h>
using namespace std;
int n;
int arr[1000005];
long long val[1000005];
long long total;
vector<pair<int, int> > v;
int front_f[1000005], back_f[1000005];
void front_inc(int i) {
  while (i != 0) {
    front_f[i]++;
    i -= (i & (-i));
  }
}
int front_query(int i) {
  int x = 0;
  while (i <= n) {
    x += front_f[i];
    i += (i & (-i));
  }
  return x;
}
void back_inc(int i) {
  while (i <= n) {
    back_f[i]++;
    i += (i & (-i));
  }
}
int back_query(int i) {
  int x = 0;
  while (i != 0) {
    x += back_f[i];
    i -= (i & (-i));
  }
  return x;
}
int main() {
  scanf("%d", &n);
  for (int x = 0; x < n; x++) {
    scanf("%d", &arr[x]);
    v.push_back(pair<int, int>(arr[x], x));
  }
  sort(v.begin(), v.end());
  for (int x = 0; x < n; x++) {
    val[x] = front_query(v[x].second + 1);
    front_inc(v[x].second + 1);
  }
  for (int x = n - 1; x >= 0; x--) {
    total += val[x] * back_query(v[x].second + 1);
    back_inc(v[x].second + 1);
  }
  printf("%lld\n", total);
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
int n;
int arr[1000005];
long long val[1000005];
long long total;
vector<pair<int, int> > v;
int front_f[1000005], back_f[1000005];
void front_inc(int i) {
  while (i != 0) {
    front_f[i]++;
    i -= (i & (-i));
  }
}
int front_query(int i) {
  int x = 0;
  while (i <= n) {
    x += front_f[i];
    i += (i & (-i));
  }
  return x;
}
void back_inc(int i) {
  while (i <= n) {
    back_f[i]++;
    i += (i & (-i));
  }
}
int back_query(int i) {
  int x = 0;
  while (i != 0) {
    x += back_f[i];
    i -= (i & (-i));
  }
  return x;
}
int main() {
  scanf("%d", &n);
  for (int x = 0; x < n; x++) {
    scanf("%d", &arr[x]);
    v.push_back(pair<int, int>(arr[x], x));
  }
  sort(v.begin(), v.end());
  for (int x = 0; x < n; x++) {
    val[x] = front_query(v[x].second + 1);
    front_inc(v[x].second + 1);
  }
  for (int x = n - 1; x >= 0; x--) {
    total += val[x] * back_query(v[x].second + 1);
    back_inc(v[x].second + 1);
  }
  printf("%lld\n", total);
}

```