#include <bits/stdc++.h>
using namespace std;
long long int seg[4000005];
long long int query(long long int x, long long int l, long long int r,
                    long long int tl, long long int tr) {
  if (tl == l && tr == r)
    return seg[x];
  else if (tl > r || tr < tl)
    return 0;
  else {
    long long int mid = (l + r) / 2;
    long long int a = 0, b = 0;
    a = query(2 * x + 1, l, mid, tl, min(tr, mid));
    b = query(2 * x + 2, mid + 1, r, max(mid + 1, tl), tr);
    return (a + b);
  }
}
int main() {
  long long int n;
  cin >> n;
  long long int ar[n];
  vector<long long int> v;
  for (long long int i = 0; i < n; i++) {
    cin >> ar[i];
    v.push_back(ar[i]);
  }
  sort(v.begin(), v.end());
  for (long long int i = 0; i < n; i++) {
    auto rank = lower_bound(v.begin(), v.end(), ar[i]) - v.begin();
    ar[i] = rank;
  }
  long long int ans = 0;
  for (long long int i = 0; i < 4000005; i++) seg[i] = 0;
  for (long long int i = 0; i < n; i++) {
    long long int rank = ar[i];
    long long int z = query(0, 0, 1000000, rank + 1, 1000000), zz;
    zz = rank - (i - z);
    ans += z * zz;
    long long int l = 0, r = 1000000, x = 0;
    while (1) {
      long long int mid = (l + r) / 2;
      seg[x]++;
      if (rank <= mid) {
        r = mid;
        x = 2 * x + 1;
      } else {
        l = mid + 1;
        x = 2 * x + 2;
      }
      if (l == r) {
        seg[x]++;
        break;
      }
    }
  }
  cout << ans << '\n';
}

### Prompt

Please create a solution in CPP 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;
long long int seg[4000005];
long long int query(long long int x, long long int l, long long int r,
                    long long int tl, long long int tr) {
  if (tl == l && tr == r)
    return seg[x];
  else if (tl > r || tr < tl)
    return 0;
  else {
    long long int mid = (l + r) / 2;
    long long int a = 0, b = 0;
    a = query(2 * x + 1, l, mid, tl, min(tr, mid));
    b = query(2 * x + 2, mid + 1, r, max(mid + 1, tl), tr);
    return (a + b);
  }
}
int main() {
  long long int n;
  cin >> n;
  long long int ar[n];
  vector<long long int> v;
  for (long long int i = 0; i < n; i++) {
    cin >> ar[i];
    v.push_back(ar[i]);
  }
  sort(v.begin(), v.end());
  for (long long int i = 0; i < n; i++) {
    auto rank = lower_bound(v.begin(), v.end(), ar[i]) - v.begin();
    ar[i] = rank;
  }
  long long int ans = 0;
  for (long long int i = 0; i < 4000005; i++) seg[i] = 0;
  for (long long int i = 0; i < n; i++) {
    long long int rank = ar[i];
    long long int z = query(0, 0, 1000000, rank + 1, 1000000), zz;
    zz = rank - (i - z);
    ans += z * zz;
    long long int l = 0, r = 1000000, x = 0;
    while (1) {
      long long int mid = (l + r) / 2;
      seg[x]++;
      if (rank <= mid) {
        r = mid;
        x = 2 * x + 1;
      } else {
        l = mid + 1;
        x = 2 * x + 2;
      }
      if (l == r) {
        seg[x]++;
        break;
      }
    }
  }
  cout << ans << '\n';
}

```