#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 1;
int arr[N];
int n, cur;
long long bit1[N];
long long bit2[N];
pair<int, int> sml[N];
void update(int i, int key, long long bit[]) {
  for (; i < N; i += (i & -i)) {
    bit[i] += key;
  }
}
long long query(int i, long long bit[]) {
  long long res = 0;
  for (; i; i -= (i & -i)) {
    res += bit[i];
  }
  return res;
}
void input() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    sml[cur++] = {arr[i], i};
  }
  sort(sml, sml + n);
  for (int i = 0, j = 1; i < n; ++i, j++) {
    arr[sml[i].second] = j;
  }
}
void solve() {
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    ans += query(N - 1, bit2) - query(arr[i], bit2);
    update(arr[i], 1, bit1);
    update(arr[i], i + 1 - query(arr[i], bit1), bit2);
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  input();
  solve();
}

### Prompt

Please formulate a cpp solution 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>
using namespace std;
const int N = 1e6 + 1;
int arr[N];
int n, cur;
long long bit1[N];
long long bit2[N];
pair<int, int> sml[N];
void update(int i, int key, long long bit[]) {
  for (; i < N; i += (i & -i)) {
    bit[i] += key;
  }
}
long long query(int i, long long bit[]) {
  long long res = 0;
  for (; i; i -= (i & -i)) {
    res += bit[i];
  }
  return res;
}
void input() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    sml[cur++] = {arr[i], i};
  }
  sort(sml, sml + n);
  for (int i = 0, j = 1; i < n; ++i, j++) {
    arr[sml[i].second] = j;
  }
}
void solve() {
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    ans += query(N - 1, bit2) - query(arr[i], bit2);
    update(arr[i], 1, bit1);
    update(arr[i], i + 1 - query(arr[i], bit1), bit2);
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  input();
  solve();
}

```