#include <bits/stdc++.h>
using namespace std;
int lrg[1000005];
int n;
void update(int indx, int val) {
  for (; indx <= n; indx += (indx & -indx)) {
    lrg[indx] += val;
  }
  return;
}
int query(int indx) {
  int sum = 0;
  for (; indx > 0; indx -= indx & -indx) sum += lrg[indx];
  return sum;
}
int main() {
  cin >> n;
  vector<pair<long long, int> > a;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    a.push_back(make_pair(x, i));
  }
  long long arr[1000001];
  sort(a.begin(), a.end());
  for (int i = 0; i < n; i++) {
    int indx = a[i].second;
    arr[indx] = i + 1;
  }
  a.clear();
  lrg[0] = 0;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    update(arr[i], 1);
    int l = query(n) - query(arr[i] - 1);
    int sml = arr[i] + l - i - 2;
    l--;
    ans += 1ll * l * sml;
  }
  cout << ans << "\n";
  return 0;
}

### 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 lrg[1000005];
int n;
void update(int indx, int val) {
  for (; indx <= n; indx += (indx & -indx)) {
    lrg[indx] += val;
  }
  return;
}
int query(int indx) {
  int sum = 0;
  for (; indx > 0; indx -= indx & -indx) sum += lrg[indx];
  return sum;
}
int main() {
  cin >> n;
  vector<pair<long long, int> > a;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    a.push_back(make_pair(x, i));
  }
  long long arr[1000001];
  sort(a.begin(), a.end());
  for (int i = 0; i < n; i++) {
    int indx = a[i].second;
    arr[indx] = i + 1;
  }
  a.clear();
  lrg[0] = 0;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    update(arr[i], 1);
    int l = query(n) - query(arr[i] - 1);
    int sml = arr[i] + l - i - 2;
    l--;
    ans += 1ll * l * sml;
  }
  cout << ans << "\n";
  return 0;
}

```