#include <bits/stdc++.h>
using namespace std;
unordered_map<long long, long long> cor;
long long n;
long long x[1000005], y[1000005];
long long perechi[1000005], unu[1000005];
void reeducare() {
  for (long long i = 1; i <= n; i++) y[i] = x[i];
  sort(y + 1, y + n + 1);
  for (long long i = 1; i <= n; i++) {
    if (y[i] != y[i - 1]) {
      cor[y[i]] = y[i - 1] + 1;
      y[i] = y[i - 1] + 1;
    } else
      y[i] = y[i - 1];
  }
  for (long long i = 1; i <= n; i++) x[i] = cor[x[i]];
}
void update(long long loc[], long long poz, long long val) {
  if (poz == 0) return;
  for (; poz <= n; poz += (poz & -poz)) loc[poz] += val;
}
long long sum(long long loc[], long long poz) {
  long long sum = 0;
  for (; poz; poz -= (poz & -poz)) sum += loc[poz];
  return sum;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> x[i];
  reeducare();
  long long rez = 0;
  for (long long i = n; i >= 1; i--) {
    rez += sum(perechi, x[i] - 1);
    update(perechi, x[i], sum(unu, x[i] - 1));
    update(unu, x[i], 1);
  }
  cout << rez;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
unordered_map<long long, long long> cor;
long long n;
long long x[1000005], y[1000005];
long long perechi[1000005], unu[1000005];
void reeducare() {
  for (long long i = 1; i <= n; i++) y[i] = x[i];
  sort(y + 1, y + n + 1);
  for (long long i = 1; i <= n; i++) {
    if (y[i] != y[i - 1]) {
      cor[y[i]] = y[i - 1] + 1;
      y[i] = y[i - 1] + 1;
    } else
      y[i] = y[i - 1];
  }
  for (long long i = 1; i <= n; i++) x[i] = cor[x[i]];
}
void update(long long loc[], long long poz, long long val) {
  if (poz == 0) return;
  for (; poz <= n; poz += (poz & -poz)) loc[poz] += val;
}
long long sum(long long loc[], long long poz) {
  long long sum = 0;
  for (; poz; poz -= (poz & -poz)) sum += loc[poz];
  return sum;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> x[i];
  reeducare();
  long long rez = 0;
  for (long long i = n; i >= 1; i--) {
    rez += sum(perechi, x[i] - 1);
    update(perechi, x[i], sum(unu, x[i] - 1));
    update(unu, x[i], 1);
  }
  cout << rez;
  return 0;
}

```