#include <bits/stdc++.h>
using namespace std;
vector<long long> bit;
long long query(long long i) {
  long long ans = 0;
  while (i > 0) {
    ans += bit[i];
    i -= i & -i;
  }
  return ans;
}
void update(long long i, long long val) {
  while (i < bit.size()) {
    bit[i] += val;
    i += i & (-i);
  }
}
long long triplets(vector<pair<long long, long long> > &arr) {
  vector<pair<long long, long long> > arr1(arr);
  sort(arr1.begin(), arr1.end());
  long long ans = 0, n = arr1.size(), before, after;
  for (long long i = n - 1; i >= 0; i--) {
    before = query(arr1[i].second);
    after = (n - arr1[i].second) - (query(n) - before);
    ans += after * before;
    update(arr1[i].second, 1);
  }
  return ans;
}
int main() {
  long long n;
  long long temp;
  cin >> n;
  bit.resize(n + 1);
  vector<pair<long long, long long> > arr;
  for (long long j = 1; j <= n; j++) {
    cin >> temp;
    arr.push_back(make_pair(temp, j));
  }
  cout << triplets(arr);
  return 0;
}

### 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;
vector<long long> bit;
long long query(long long i) {
  long long ans = 0;
  while (i > 0) {
    ans += bit[i];
    i -= i & -i;
  }
  return ans;
}
void update(long long i, long long val) {
  while (i < bit.size()) {
    bit[i] += val;
    i += i & (-i);
  }
}
long long triplets(vector<pair<long long, long long> > &arr) {
  vector<pair<long long, long long> > arr1(arr);
  sort(arr1.begin(), arr1.end());
  long long ans = 0, n = arr1.size(), before, after;
  for (long long i = n - 1; i >= 0; i--) {
    before = query(arr1[i].second);
    after = (n - arr1[i].second) - (query(n) - before);
    ans += after * before;
    update(arr1[i].second, 1);
  }
  return ans;
}
int main() {
  long long n;
  long long temp;
  cin >> n;
  bit.resize(n + 1);
  vector<pair<long long, long long> > arr;
  for (long long j = 1; j <= n; j++) {
    cin >> temp;
    arr.push_back(make_pair(temp, j));
  }
  cout << triplets(arr);
  return 0;
}

```