#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int n;
vector<ll> ft;
void add(ll i, ll d) {
  for (; i <= n; i += i & -i) ft[i] += d;
}
ll sum(ll i) {
  if (i == 0) return 0;
  return sum(i ^ (i & -i)) + ft[i];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<int> a(n + 1);
  vector<int> b(n + 1);
  ft.assign(n + 1, 0);
  vector<pair<int, int>> asd(n + 1, make_pair(0, 0));
  for (int i = 1; i < n + 1; i++) {
    cin >> asd[i].first;
    asd[i].second = i;
  }
  sort(asd.begin(), asd.end());
  vector<int> dsa(n + 1);
  int j = 1;
  dsa[1] = asd[1].first = 1;
  for (int i = 2; i < n + 1; i++) {
    if (asd[i - 1].first != asd[i].first) j++;
    dsa[i] = j;
  }
  for (int i = 1; i < n + 1; i++) {
    a[asd[i].second] = dsa[i];
  }
  for (int i = 1; i < n + 1; i++) {
    b[i] = i - sum(a[i]) - 1;
    add(a[i], 1);
  }
  ft.assign(n + 1, 0);
  ll s = 0;
  for (int i = 1; i < n + 1; i++) {
    add(a[i], b[i]);
    s += sum(n) - sum(a[i]);
  }
  cout << s << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
using ll = long long;
int n;
vector<ll> ft;
void add(ll i, ll d) {
  for (; i <= n; i += i & -i) ft[i] += d;
}
ll sum(ll i) {
  if (i == 0) return 0;
  return sum(i ^ (i & -i)) + ft[i];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<int> a(n + 1);
  vector<int> b(n + 1);
  ft.assign(n + 1, 0);
  vector<pair<int, int>> asd(n + 1, make_pair(0, 0));
  for (int i = 1; i < n + 1; i++) {
    cin >> asd[i].first;
    asd[i].second = i;
  }
  sort(asd.begin(), asd.end());
  vector<int> dsa(n + 1);
  int j = 1;
  dsa[1] = asd[1].first = 1;
  for (int i = 2; i < n + 1; i++) {
    if (asd[i - 1].first != asd[i].first) j++;
    dsa[i] = j;
  }
  for (int i = 1; i < n + 1; i++) {
    a[asd[i].second] = dsa[i];
  }
  for (int i = 1; i < n + 1; i++) {
    b[i] = i - sum(a[i]) - 1;
    add(a[i], 1);
  }
  ft.assign(n + 1, 0);
  ll s = 0;
  for (int i = 1; i < n + 1; i++) {
    add(a[i], b[i]);
    s += sum(n) - sum(a[i]);
  }
  cout << s << endl;
  return 0;
}

```