#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
long long fen[2][maxn];
pair<int, int> a[maxn];
void add(int type, int idx, int val) {
  for (; idx < maxn; idx += idx & -idx) fen[type][idx] += val;
}
long long get(int type, int idx) {
  long long ret = 0;
  for (; idx > 0; idx -= idx & -idx) ret += fen[type][idx];
  return ret;
}
bool cmp(pair<int, int> b, pair<int, int> c) {
  if (b.second < c.second) return true;
  return false;
}
int main() {
  long long ans = 0;
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a + 1, a + n + 1);
  a[1].first = 1;
  add(1, 1, 1);
  for (int i = 2; i <= n; i++) {
    if (a[i].first == a[i - 1].first)
      a[i].first = a[i - 1].first;
    else
      a[i].first = a[i - 1].first + 1;
    add(1, a[i].first, 1);
  }
  sort(a + 1, a + n + 1, cmp);
  add(1, a[1].first, -1);
  add(0, a[1].first, 1);
  for (int i = 2; i < n; i++) {
    ans += get(1, a[i].first - 1) * (get(0, maxn - 1) - get(0, a[i].first));
    add(1, a[i].first, -1);
    add(0, a[i].first, 1);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Generate 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 maxn = 1e6 + 10;
long long fen[2][maxn];
pair<int, int> a[maxn];
void add(int type, int idx, int val) {
  for (; idx < maxn; idx += idx & -idx) fen[type][idx] += val;
}
long long get(int type, int idx) {
  long long ret = 0;
  for (; idx > 0; idx -= idx & -idx) ret += fen[type][idx];
  return ret;
}
bool cmp(pair<int, int> b, pair<int, int> c) {
  if (b.second < c.second) return true;
  return false;
}
int main() {
  long long ans = 0;
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a + 1, a + n + 1);
  a[1].first = 1;
  add(1, 1, 1);
  for (int i = 2; i <= n; i++) {
    if (a[i].first == a[i - 1].first)
      a[i].first = a[i - 1].first;
    else
      a[i].first = a[i - 1].first + 1;
    add(1, a[i].first, 1);
  }
  sort(a + 1, a + n + 1, cmp);
  add(1, a[1].first, -1);
  add(0, a[1].first, 1);
  for (int i = 2; i < n; i++) {
    ans += get(1, a[i].first - 1) * (get(0, maxn - 1) - get(0, a[i].first));
    add(1, a[i].first, -1);
    add(0, a[i].first, 1);
  }
  cout << ans << endl;
  return 0;
}

```