#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9;
const long long MOD = 1e9 + 7;
const int MAXN = 1e6 + 100;
int A[MAXN];
int num = 1;
map<int, int> mep;
map<int, int>::iterator it;
long long ft0[MAXN], ft1[MAXN];
int n;
void update(int idx, int val, bool flag) {
  for (; idx < MAXN; idx += (idx & (-idx))) {
    if (flag)
      ft1[idx] += val;
    else
      ft0[idx] += val;
  }
}
long long query(int idx, bool flag) {
  long long res = 0;
  for (; idx; idx -= (idx & (-idx))) {
    if (flag)
      res += ft1[idx];
    else
      res += ft0[idx];
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    mep[A[i]];
  }
  for (it = mep.begin(); it != mep.end(); it++) {
    mep[it->first] = num;
    update(num++, 1, 1);
  }
  long long tot = 0;
  for (int i = 1; i <= n; i++) {
    int idx = mep[A[i]];
    update(idx, -1, 1);
    tot += ((query(num - 1, 0) - query(idx, 0)) * query(idx - 1, 1));
    update(idx, 1, 0);
  }
  cout << tot << '\n';
}

### Prompt

Develop a solution in CPP 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 long long INF = 1e9;
const long long MOD = 1e9 + 7;
const int MAXN = 1e6 + 100;
int A[MAXN];
int num = 1;
map<int, int> mep;
map<int, int>::iterator it;
long long ft0[MAXN], ft1[MAXN];
int n;
void update(int idx, int val, bool flag) {
  for (; idx < MAXN; idx += (idx & (-idx))) {
    if (flag)
      ft1[idx] += val;
    else
      ft0[idx] += val;
  }
}
long long query(int idx, bool flag) {
  long long res = 0;
  for (; idx; idx -= (idx & (-idx))) {
    if (flag)
      res += ft1[idx];
    else
      res += ft0[idx];
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    mep[A[i]];
  }
  for (it = mep.begin(); it != mep.end(); it++) {
    mep[it->first] = num;
    update(num++, 1, 1);
  }
  long long tot = 0;
  for (int i = 1; i <= n; i++) {
    int idx = mep[A[i]];
    update(idx, -1, 1);
    tot += ((query(num - 1, 0) - query(idx, 0)) * query(idx - 1, 1));
    update(idx, 1, 0);
  }
  cout << tot << '\n';
}

```