#include <bits/stdc++.h>
using namespace std;
const int maxn = 3 * 1e6;
int n, ft[2][maxn], id, A[maxn], F[2][maxn];
set<pair<int, int> > S;
void update(int k, int i) {
  for (; k < n; k += (k & (-k))) ft[i][k]++;
}
int query(int a, int i) {
  int res = 0;
  for (; a; a -= (a & -a)) res += ft[i][a];
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> A[i], S.emplace(A[i], i);
  for (auto p : S) A[p.second] = ++id;
  for (int k = 0; k < 2; ++k) {
    reverse(A, A + n);
    for (int i = 0; i < n; ++i) {
      F[k][i] = query(A[i], k);
      update(A[i], k);
      A[i] = n - A[i] + 1;
    }
  }
  long long res = 0;
  for (int i = 0; i < n; ++i) res += F[0][i] * 1LL * F[1][n - i - 1];
  cout << res << '\n';
}

### 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;
const int maxn = 3 * 1e6;
int n, ft[2][maxn], id, A[maxn], F[2][maxn];
set<pair<int, int> > S;
void update(int k, int i) {
  for (; k < n; k += (k & (-k))) ft[i][k]++;
}
int query(int a, int i) {
  int res = 0;
  for (; a; a -= (a & -a)) res += ft[i][a];
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> A[i], S.emplace(A[i], i);
  for (auto p : S) A[p.second] = ++id;
  for (int k = 0; k < 2; ++k) {
    reverse(A, A + n);
    for (int i = 0; i < n; ++i) {
      F[k][i] = query(A[i], k);
      update(A[i], k);
      A[i] = n - A[i] + 1;
    }
  }
  long long res = 0;
  for (int i = 0; i < n; ++i) res += F[0][i] * 1LL * F[1][n - i - 1];
  cout << res << '\n';
}

```