#include <bits/stdc++.h>
using namespace std;
struct BITree {
  vector<long long> values;
  int size;
  BITree(int n) {
    size = n;
    values.resize(size + 1);
  }
  long long getSum(int i) {
    i++;
    long long sum = 0;
    while (i <= size) {
      sum += values[i];
      i += i & -i;
    }
    return sum;
  }
  void update(int i, long long v) {
    i++;
    while (i > 0) {
      values[i] += v;
      i -= i & -i;
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  cin >> n;
  vector<pair<int, int>> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a.begin(), a.end());
  vector<int> power(n);
  for (int i = 0; i < n; i++) power[a[i].second] = i;
  vector<int> doubles(n);
  BITree tree(n);
  for (int i = 0; i < n; i++) {
    tree.update(power[i] - 1, 1);
    doubles[power[i]] = tree.getSum(power[i]);
  }
  tree = BITree(n);
  vector<long long> triples(n);
  for (int i = 0; i < n; i++) {
    tree.update(power[i] - 1, doubles[power[i]]);
    triples[power[i]] = tree.getSum(power[i]);
  }
  long long ans = 0;
  for (long long t : triples) ans += t;
  cout << ans;
}

### Prompt

Please provide a CPP coded solution 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;
struct BITree {
  vector<long long> values;
  int size;
  BITree(int n) {
    size = n;
    values.resize(size + 1);
  }
  long long getSum(int i) {
    i++;
    long long sum = 0;
    while (i <= size) {
      sum += values[i];
      i += i & -i;
    }
    return sum;
  }
  void update(int i, long long v) {
    i++;
    while (i > 0) {
      values[i] += v;
      i -= i & -i;
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  cin >> n;
  vector<pair<int, int>> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a.begin(), a.end());
  vector<int> power(n);
  for (int i = 0; i < n; i++) power[a[i].second] = i;
  vector<int> doubles(n);
  BITree tree(n);
  for (int i = 0; i < n; i++) {
    tree.update(power[i] - 1, 1);
    doubles[power[i]] = tree.getSum(power[i]);
  }
  tree = BITree(n);
  vector<long long> triples(n);
  for (int i = 0; i < n; i++) {
    tree.update(power[i] - 1, doubles[power[i]]);
    triples[power[i]] = tree.getSum(power[i]);
  }
  long long ans = 0;
  for (long long t : triples) ans += t;
  cout << ans;
}

```