#include <bits/stdc++.h>
using namespace std;
int n, a[1000001];
pair<int, int> p[1000001];
long long ans, Left[1000001], Right[1000001], bit[1000001];
void upd(int ind, int x) {
  while (ind <= n) {
    bit[ind] += x;
    ind += ind & (-ind);
  }
}
long long get(int ind) {
  long long ans = 0;
  while (ind > 0) {
    ans += bit[ind];
    ind -= ind & (-ind);
  }
  return ans;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i], p[i].first = a[i], p[i].second = i;
  sort(p + 1, p + n + 1);
  for (int i = 1; i <= n; i++) {
    a[p[i].second] = i;
  }
  for (int i = 1; i <= n; i++) {
    Right[i] = get(n) - get(a[i]);
    upd(a[i], 1);
  }
  for (int i = 1; i <= n; i++) upd(a[i], -1);
  for (int i = n; i > 0; i--) {
    Left[i] = get(a[i] - 1);
    upd(a[i], 1);
  }
  for (int i = 1; i <= n; i++) ans += Left[i] * Right[i];
  cout << ans;
}

### Prompt

Please formulate 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;
int n, a[1000001];
pair<int, int> p[1000001];
long long ans, Left[1000001], Right[1000001], bit[1000001];
void upd(int ind, int x) {
  while (ind <= n) {
    bit[ind] += x;
    ind += ind & (-ind);
  }
}
long long get(int ind) {
  long long ans = 0;
  while (ind > 0) {
    ans += bit[ind];
    ind -= ind & (-ind);
  }
  return ans;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i], p[i].first = a[i], p[i].second = i;
  sort(p + 1, p + n + 1);
  for (int i = 1; i <= n; i++) {
    a[p[i].second] = i;
  }
  for (int i = 1; i <= n; i++) {
    Right[i] = get(n) - get(a[i]);
    upd(a[i], 1);
  }
  for (int i = 1; i <= n; i++) upd(a[i], -1);
  for (int i = n; i > 0; i--) {
    Left[i] = get(a[i] - 1);
    upd(a[i], 1);
  }
  for (int i = 1; i <= n; i++) ans += Left[i] * Right[i];
  cout << ans;
}

```