#include <bits/stdc++.h>
using namespace std;
const int Max = 2e6 + 10;
int A[Max];
int cnv[Max][2];
int bit[Max];
int m;
void ins(int x) {
  while (x <= m) bit[x]++, x += x & -x;
}
int get(int x) {
  int ans = 0;
  while (x >= 1) ans += bit[x], x -= x & -x;
  return ans;
}
int bigback[Max];
int smallfore[Max];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int> > used;
  for (int i = 0; i < n; i++) {
    cin >> A[i];
    used.push_back(make_pair(A[i], 2 * i));
    used.push_back(make_pair(A[i] - 1, 2 * i + 1));
  }
  sort(used.begin(), used.end());
  for (int i = 0, j = 1; i < used.size(); i++)
    cnv[used[i].second / 2][used[i].second % 2] = j =
        j + (i > 0 && used[i].first != used[i - 1].first);
  used.resize(unique(used.begin(), used.end()) - used.begin());
  m = used.size();
  for (int i = 0; i < n; i++) {
    bigback[i] = i - get(cnv[i][0]);
    ins(cnv[i][0]);
  }
  for (int i = 1; i <= m; i++) bit[i] = 0;
  for (int i = n; i--;) {
    smallfore[i] = get(cnv[i][1]);
    ins(cnv[i][0]);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) ans += 1ll * bigback[i] * smallfore[i];
  cout << ans << '\n';
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
const int Max = 2e6 + 10;
int A[Max];
int cnv[Max][2];
int bit[Max];
int m;
void ins(int x) {
  while (x <= m) bit[x]++, x += x & -x;
}
int get(int x) {
  int ans = 0;
  while (x >= 1) ans += bit[x], x -= x & -x;
  return ans;
}
int bigback[Max];
int smallfore[Max];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int> > used;
  for (int i = 0; i < n; i++) {
    cin >> A[i];
    used.push_back(make_pair(A[i], 2 * i));
    used.push_back(make_pair(A[i] - 1, 2 * i + 1));
  }
  sort(used.begin(), used.end());
  for (int i = 0, j = 1; i < used.size(); i++)
    cnv[used[i].second / 2][used[i].second % 2] = j =
        j + (i > 0 && used[i].first != used[i - 1].first);
  used.resize(unique(used.begin(), used.end()) - used.begin());
  m = used.size();
  for (int i = 0; i < n; i++) {
    bigback[i] = i - get(cnv[i][0]);
    ins(cnv[i][0]);
  }
  for (int i = 1; i <= m; i++) bit[i] = 0;
  for (int i = n; i--;) {
    smallfore[i] = get(cnv[i][1]);
    ins(cnv[i][0]);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) ans += 1ll * bigback[i] * smallfore[i];
  cout << ans << '\n';
}

```