#include <bits/stdc++.h>
using namespace std;
const int N = 1e6;
long long fn[2][N + 5], temp[N + 5], v[N + 5];
int n;
long long query(long long tree[], int k) {
  long long s = 0;
  while (k > 0) {
    s += tree[k];
    k -= k & -k;
  }
  return s;
}
void update(long long tree[], int k, int val) {
  while (k <= n) {
    tree[k] += val;
    k += k & -k;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long s = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> temp[i];
    v[i] = temp[i];
  }
  sort(temp + 1, temp + n + 1);
  for (int i = 1; i <= n; i++) {
    int poz = lower_bound(temp + 1, temp + n + 1, v[i]) - temp;
    v[i] = poz;
  }
  for (int i = 1; i <= n; i++) {
    s += query(fn[1], n) - query(fn[1], v[i]);
    update(fn[0], v[i], 1);
    update(fn[1], v[i], query(fn[0], n) - query(fn[0], v[i]));
  }
  cout << s;
}

### Prompt

Your task is to create 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;
long long fn[2][N + 5], temp[N + 5], v[N + 5];
int n;
long long query(long long tree[], int k) {
  long long s = 0;
  while (k > 0) {
    s += tree[k];
    k -= k & -k;
  }
  return s;
}
void update(long long tree[], int k, int val) {
  while (k <= n) {
    tree[k] += val;
    k += k & -k;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long s = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> temp[i];
    v[i] = temp[i];
  }
  sort(temp + 1, temp + n + 1);
  for (int i = 1; i <= n; i++) {
    int poz = lower_bound(temp + 1, temp + n + 1, v[i]) - temp;
    v[i] = poz;
  }
  for (int i = 1; i <= n; i++) {
    s += query(fn[1], n) - query(fn[1], v[i]);
    update(fn[0], v[i], 1);
    update(fn[1], v[i], query(fn[0], n) - query(fn[0], v[i]));
  }
  cout << s;
}

```