#include <bits/stdc++.h>
using namespace std;
const int maxN = 1e6 + 3;
int n;
vector<int> fenf(maxN), fenr(maxN);
void add(int l, int d, vector<int> &fen) {
  while (l < n) {
    fen[l] += d;
    l = (l | (l + 1));
  }
}
int sum(int r, vector<int> &fen) {
  int ans = 0;
  while (r >= 0) {
    ans += fen[r];
    r = (r & (r + 1)) - 1;
  }
  return ans;
}
int range_sum(int l, int r, vector<int> &fen) {
  int ans = sum(r, fen) - sum(l, fen);
  return ans;
}
void solve() {
  cin >> n;
  vector<int> a(n), h(n);
  for (int i = 0; i < n; i++) {
    cin >> h[i];
    a[i] = h[i];
  }
  sort(h.begin(), h.end());
  for (int i = 0; i < n; i++) {
    a[i] = lower_bound(h.begin(), h.end(), a[i]) - h.begin();
  }
  for (int i = 0; i < n; i++) {
    add(a[i], 1, fenr);
  }
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    add(a[i], -1, fenr);
    long long int left = range_sum(a[i], n - 1, fenf);
    long long int right = sum(a[i] - 1, fenr);
    ans += left * right;
    add(a[i], 1, fenf);
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q = 1;
  while (q--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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 maxN = 1e6 + 3;
int n;
vector<int> fenf(maxN), fenr(maxN);
void add(int l, int d, vector<int> &fen) {
  while (l < n) {
    fen[l] += d;
    l = (l | (l + 1));
  }
}
int sum(int r, vector<int> &fen) {
  int ans = 0;
  while (r >= 0) {
    ans += fen[r];
    r = (r & (r + 1)) - 1;
  }
  return ans;
}
int range_sum(int l, int r, vector<int> &fen) {
  int ans = sum(r, fen) - sum(l, fen);
  return ans;
}
void solve() {
  cin >> n;
  vector<int> a(n), h(n);
  for (int i = 0; i < n; i++) {
    cin >> h[i];
    a[i] = h[i];
  }
  sort(h.begin(), h.end());
  for (int i = 0; i < n; i++) {
    a[i] = lower_bound(h.begin(), h.end(), a[i]) - h.begin();
  }
  for (int i = 0; i < n; i++) {
    add(a[i], 1, fenr);
  }
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    add(a[i], -1, fenr);
    long long int left = range_sum(a[i], n - 1, fenf);
    long long int right = sum(a[i] - 1, fenr);
    ans += left * right;
    add(a[i], 1, fenf);
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q = 1;
  while (q--) {
    solve();
  }
  return 0;
}

```