#include <bits/stdc++.h>
using namespace std;
inline void fen_update(vector<int> &tree, int i, int val) {
  for (int k = i; k > 0 && k < tree.size(); k += (k & -k)) tree[k] += val;
}
inline int fen_query(vector<int> &tree, int i) {
  int sum = 0;
  for (int k = i; k > 0; k &= k - 1) sum += tree[k];
  return sum;
}
int main(int argc, char *argv[]) {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<pair<int, int> > t;
  for (int i = 0; i < n; i++) cin >> a[i], t.push_back(make_pair(a[i], i));
  sort(t.begin(), t.end());
  for (int i = 0, j = 0; i < n; i++) {
    if (i > 0 && t[i].first != t[i - 1].first) j++;
    a[t[i].second] = j + 1;
  }
  vector<int> fen1(n + 1), fen2(n + 1);
  for (int i = 0; i < n; i++)
    for (int k = a[i]; k <= n; k += (k & -k)) fen2[k]++;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    fen_update(fen2, a[i], -1);
    long long c1 = fen_query(fen1, n) - fen_query(fen1, a[i]);
    long long c2 = fen_query(fen2, a[i] - 1);
    ans += c1 * c2;
    fen_update(fen1, a[i], 1);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
inline void fen_update(vector<int> &tree, int i, int val) {
  for (int k = i; k > 0 && k < tree.size(); k += (k & -k)) tree[k] += val;
}
inline int fen_query(vector<int> &tree, int i) {
  int sum = 0;
  for (int k = i; k > 0; k &= k - 1) sum += tree[k];
  return sum;
}
int main(int argc, char *argv[]) {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<pair<int, int> > t;
  for (int i = 0; i < n; i++) cin >> a[i], t.push_back(make_pair(a[i], i));
  sort(t.begin(), t.end());
  for (int i = 0, j = 0; i < n; i++) {
    if (i > 0 && t[i].first != t[i - 1].first) j++;
    a[t[i].second] = j + 1;
  }
  vector<int> fen1(n + 1), fen2(n + 1);
  for (int i = 0; i < n; i++)
    for (int k = a[i]; k <= n; k += (k & -k)) fen2[k]++;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    fen_update(fen2, a[i], -1);
    long long c1 = fen_query(fen1, n) - fen_query(fen1, a[i]);
    long long c2 = fen_query(fen2, a[i] - 1);
    ans += c1 * c2;
    fen_update(fen1, a[i], 1);
  }
  cout << ans << endl;
  return 0;
}

```