#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 20;
long long sum;
int k, p, n, a[N + 1], i, b[N + 1], left_tree[N + 1], right_tree[N + 1];
vector<long long> leftv, rightv;
void update(int index, int value, int tree[]) {
  while (index < N + 1) {
    tree[index] += value;
    index += (index & -index);
  }
}
int read(int index, int tree[]) {
  int sum = 0;
  while (index > 0) {
    sum += tree[index];
    index -= (index & -index);
  }
  return sum;
}
int main() {
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  for (i = 0; i < n; i++) {
    a[i] = upper_bound(b, b + n, a[i]) - b;
  }
  for (i = 0; i < n; i++) {
    update(a[i], 1, left_tree);
    leftv.push_back(read(N, left_tree) - read(a[i], left_tree));
  }
  for (i = n - 1; i >= 0; i--) {
    rightv.push_back(read(a[i], right_tree));
    update(a[i], 1, right_tree);
  }
  reverse(rightv.begin(), rightv.end());
  for (i = 0; i < n; i++) {
    sum += rightv[i] * leftv[i];
  }
  cout << sum;
}

### Prompt

In cpp, your task is to solve 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;
const int N = 1 << 20;
long long sum;
int k, p, n, a[N + 1], i, b[N + 1], left_tree[N + 1], right_tree[N + 1];
vector<long long> leftv, rightv;
void update(int index, int value, int tree[]) {
  while (index < N + 1) {
    tree[index] += value;
    index += (index & -index);
  }
}
int read(int index, int tree[]) {
  int sum = 0;
  while (index > 0) {
    sum += tree[index];
    index -= (index & -index);
  }
  return sum;
}
int main() {
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  for (i = 0; i < n; i++) {
    a[i] = upper_bound(b, b + n, a[i]) - b;
  }
  for (i = 0; i < n; i++) {
    update(a[i], 1, left_tree);
    leftv.push_back(read(N, left_tree) - read(a[i], left_tree));
  }
  for (i = n - 1; i >= 0; i--) {
    rightv.push_back(read(a[i], right_tree));
    update(a[i], 1, right_tree);
  }
  reverse(rightv.begin(), rightv.end());
  for (i = 0; i < n; i++) {
    sum += rightv[i] * leftv[i];
  }
  cout << sum;
}

```