#include <bits/stdc++.h>
using namespace std;
long long int bit[1000010] = {0};
long long int get_sum(int i) {
  long long int ans = 0;
  while (i > 0) {
    ans += bit[i];
    i -= i & (-i);
  }
  return ans;
}
void update(int i, int n) {
  while (i <= n) {
    bit[i] += 1;
    i += i & (-i);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  long long int arr[n], small[n], big[n], tmp[n];
  long long int max1 = 0;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    tmp[i] = arr[i];
  }
  sort(tmp, tmp + n);
  for (int i = 0; i < n; i++) {
    arr[i] = lower_bound(tmp, tmp + n, arr[i]) - tmp + 1;
    if (arr[i] > max1) max1 = arr[i];
  }
  long long int ans = 0;
  for (int i = (n - 1); i >= 0; i--) {
    small[i] = get_sum(arr[i] - 1);
    update(arr[i], max1);
  }
  for (int i = 0; i <= max1; i++) bit[i] = 0;
  for (int i = 0; i < n; i++) {
    big[i] = i - get_sum(arr[i]);
    update(arr[i], max1);
    ans += big[i] * small[i];
  }
  cout << ans;
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
long long int bit[1000010] = {0};
long long int get_sum(int i) {
  long long int ans = 0;
  while (i > 0) {
    ans += bit[i];
    i -= i & (-i);
  }
  return ans;
}
void update(int i, int n) {
  while (i <= n) {
    bit[i] += 1;
    i += i & (-i);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  long long int arr[n], small[n], big[n], tmp[n];
  long long int max1 = 0;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    tmp[i] = arr[i];
  }
  sort(tmp, tmp + n);
  for (int i = 0; i < n; i++) {
    arr[i] = lower_bound(tmp, tmp + n, arr[i]) - tmp + 1;
    if (arr[i] > max1) max1 = arr[i];
  }
  long long int ans = 0;
  for (int i = (n - 1); i >= 0; i--) {
    small[i] = get_sum(arr[i] - 1);
    update(arr[i], max1);
  }
  for (int i = 0; i <= max1; i++) bit[i] = 0;
  for (int i = 0; i < n; i++) {
    big[i] = i - get_sum(arr[i]);
    update(arr[i], max1);
    ans += big[i] * small[i];
  }
  cout << ans;
  return 0;
}

```