#include <bits/stdc++.h>
using namespace std;
const long long N = 1e6 + 5;
long long n, a[N], f[N], ans, f1[N], f2[N];
void add(long long x, long long c) {
  for (; x < N; x += (x & -x)) f[x] += c;
}
long long get(long long x) {
  long long cnt = 0;
  for (; 0 < x; x -= (x & -x)) cnt += f[x];
  return cnt;
}
void compress() {
  vector<long long> u;
  for (int i = 0; i < n; i++) u.push_back(a[i]);
  sort(u.begin(), u.end());
  u.resize(unique(u.begin(), u.end()) - u.begin());
  for (int i = 0; i < n; i++)
    a[i] = lower_bound(u.begin(), u.end(), a[i]) - u.begin();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  compress();
  for (int i = 0; i < n; i++) a[i]++;
  for (int i = 0; i < n; i++) {
    add(a[i], 1);
    f1[i] = i - get(a[i]) + 1;
  }
  for (int i = 0; i < n; i++) {
    add(a[i], -1);
  }
  for (int i = n - 1; i >= 0; i--) {
    add(a[i], 1);
    f2[i] = get(a[i] - 1);
  }
  for (int i = 0; i < n; i++) {
    ans += f1[i] * f2[i];
  }
  cout << ans;
}

### Prompt

Generate 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 long long N = 1e6 + 5;
long long n, a[N], f[N], ans, f1[N], f2[N];
void add(long long x, long long c) {
  for (; x < N; x += (x & -x)) f[x] += c;
}
long long get(long long x) {
  long long cnt = 0;
  for (; 0 < x; x -= (x & -x)) cnt += f[x];
  return cnt;
}
void compress() {
  vector<long long> u;
  for (int i = 0; i < n; i++) u.push_back(a[i]);
  sort(u.begin(), u.end());
  u.resize(unique(u.begin(), u.end()) - u.begin());
  for (int i = 0; i < n; i++)
    a[i] = lower_bound(u.begin(), u.end(), a[i]) - u.begin();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  compress();
  for (int i = 0; i < n; i++) a[i]++;
  for (int i = 0; i < n; i++) {
    add(a[i], 1);
    f1[i] = i - get(a[i]) + 1;
  }
  for (int i = 0; i < n; i++) {
    add(a[i], -1);
  }
  for (int i = n - 1; i >= 0; i--) {
    add(a[i], 1);
    f2[i] = get(a[i] - 1);
  }
  for (int i = 0; i < n; i++) {
    ans += f1[i] * f2[i];
  }
  cout << ans;
}

```