#include <bits/stdc++.h>
using namespace std;
int a[1000100], n;
vector<long long> v;
long long fen[1000010];
long long fen2[1000010];
void add(int x) {
  for (int i = x; i > 0; i -= (i & -i)) {
    fen[i]++;
  }
}
void add1(int x, long long k) {
  for (int i = x; i > 0; i -= (i & -i)) {
    fen2[i] += k;
  }
}
long long fine1(int x) {
  long long res = 0;
  for (int i = x; i <= n + 1; i += (i & -i)) {
    res += fen2[i];
  }
  return res;
}
long long fine(int x) {
  long long res = 0;
  for (int i = x; i <= n + 1; i += (i & -i)) {
    res += fen[i];
  }
  return res;
}
int main() {
  long long ans = 0;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  for (int i = 0; i < n; ++i) {
    a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
    add(a[i]);
    long long u = fine(a[i] + 1);
    if (u) {
      add1(a[i], u);
    }
    ans += fine1(a[i] + 1);
  }
  cout << ans;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int a[1000100], n;
vector<long long> v;
long long fen[1000010];
long long fen2[1000010];
void add(int x) {
  for (int i = x; i > 0; i -= (i & -i)) {
    fen[i]++;
  }
}
void add1(int x, long long k) {
  for (int i = x; i > 0; i -= (i & -i)) {
    fen2[i] += k;
  }
}
long long fine1(int x) {
  long long res = 0;
  for (int i = x; i <= n + 1; i += (i & -i)) {
    res += fen2[i];
  }
  return res;
}
long long fine(int x) {
  long long res = 0;
  for (int i = x; i <= n + 1; i += (i & -i)) {
    res += fen[i];
  }
  return res;
}
int main() {
  long long ans = 0;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  for (int i = 0; i < n; ++i) {
    a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
    add(a[i]);
    long long u = fine(a[i] + 1);
    if (u) {
      add1(a[i], u);
    }
    ans += fine1(a[i] + 1);
  }
  cout << ans;
  return 0;
}

```