#include <bits/stdc++.h>
using namespace std;
const int N = 1000 * 1000 + 100;
struct Fenwick {
  int fen[N];
  void add(int p, int val) {
    for (p++; p <= N; p += p & -p) fen[p] += val;
  }
  long long get(int p, int sum = 0) {
    for (; p; p -= p & -p) sum += fen[p];
    return sum;
  }
  Fenwick() { memset(fen, 0, sizeof fen); }
};
Fenwick f[2];
int n, a[N], b[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  int sz = unique(b, b + n) - b;
  for (int i = 0; i < n; ++i) {
    a[i] = lower_bound(b, b + sz, a[i]) - b;
    f[1].add(a[i], 1);
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    f[1].add(a[i], -1);
    ans += (f[0].get(N) - f[0].get(a[i] + 1)) * f[1].get(a[i]);
    f[0].add(a[i], 1);
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 = 1000 * 1000 + 100;
struct Fenwick {
  int fen[N];
  void add(int p, int val) {
    for (p++; p <= N; p += p & -p) fen[p] += val;
  }
  long long get(int p, int sum = 0) {
    for (; p; p -= p & -p) sum += fen[p];
    return sum;
  }
  Fenwick() { memset(fen, 0, sizeof fen); }
};
Fenwick f[2];
int n, a[N], b[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  int sz = unique(b, b + n) - b;
  for (int i = 0; i < n; ++i) {
    a[i] = lower_bound(b, b + sz, a[i]) - b;
    f[1].add(a[i], 1);
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    f[1].add(a[i], -1);
    ans += (f[0].get(N) - f[0].get(a[i] + 1)) * f[1].get(a[i]);
    f[0].add(a[i], 1);
  }
  cout << ans << "\n";
  return 0;
}

```