#include <bits/stdc++.h>
using namespace std;
const long long int N = 1e6 + 5;
const long long int mod = 1e9 + 7;
const long long int level = 20;
vector<pair<long long int, long long int> > v;
long long int tree[4 * N];
long long int query(long long int n, long long int s, long long int e,
                    long long int l, long long int r) {
  if (r < s or l > e) return 0;
  if (l <= s and r >= e) return tree[n];
  long long int m = (s + e) >> 1;
  return query(2 * n, s, m, l, r) + query(2 * n + 1, m + 1, e, l, r);
}
void update(long long int n, long long int s, long long int e,
            long long int idx) {
  if (idx < s or idx > e) return;
  if (s == e) {
    tree[n] = 1;
    return;
  }
  long long int m = (s + e) >> 1;
  update(2 * n, s, m, idx);
  update(2 * n + 1, m + 1, e, idx);
  tree[n] = tree[2 * n] + tree[2 * n + 1];
}
int main() {
  ios::sync_with_stdio(false);
  long long int n;
  cin >> n;
  long long int arr[n + 1];
  long long int ltr[n + 1];
  long long int rtl[n + 1];
  for (long long int i = 1; i < n + 1; i++) {
    long long int x;
    cin >> x;
    v.push_back({x, i});
  }
  sort(v.begin(), v.end());
  for (long long int i = 0; i < n; i++) {
    arr[v[i].second] = i + 1;
  }
  long long int ans = 0;
  for (long long int i = n + 1 - 1; i >= 1; i--) {
    long long int x = query(1, 1, n, 1, arr[i]);
    rtl[i] = x;
    update(1, 1, n, arr[i]);
  }
  memset(tree, 0, sizeof(tree));
  for (long long int i = 1; i < n + 1; i++) {
    long long int x = query(1, 1, n, arr[i], n);
    ltr[i] = x;
    update(1, 1, n, arr[i]);
    ans += rtl[i] * ltr[i];
  }
  cout << ans << endl;
  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;
const long long int N = 1e6 + 5;
const long long int mod = 1e9 + 7;
const long long int level = 20;
vector<pair<long long int, long long int> > v;
long long int tree[4 * N];
long long int query(long long int n, long long int s, long long int e,
                    long long int l, long long int r) {
  if (r < s or l > e) return 0;
  if (l <= s and r >= e) return tree[n];
  long long int m = (s + e) >> 1;
  return query(2 * n, s, m, l, r) + query(2 * n + 1, m + 1, e, l, r);
}
void update(long long int n, long long int s, long long int e,
            long long int idx) {
  if (idx < s or idx > e) return;
  if (s == e) {
    tree[n] = 1;
    return;
  }
  long long int m = (s + e) >> 1;
  update(2 * n, s, m, idx);
  update(2 * n + 1, m + 1, e, idx);
  tree[n] = tree[2 * n] + tree[2 * n + 1];
}
int main() {
  ios::sync_with_stdio(false);
  long long int n;
  cin >> n;
  long long int arr[n + 1];
  long long int ltr[n + 1];
  long long int rtl[n + 1];
  for (long long int i = 1; i < n + 1; i++) {
    long long int x;
    cin >> x;
    v.push_back({x, i});
  }
  sort(v.begin(), v.end());
  for (long long int i = 0; i < n; i++) {
    arr[v[i].second] = i + 1;
  }
  long long int ans = 0;
  for (long long int i = n + 1 - 1; i >= 1; i--) {
    long long int x = query(1, 1, n, 1, arr[i]);
    rtl[i] = x;
    update(1, 1, n, arr[i]);
  }
  memset(tree, 0, sizeof(tree));
  for (long long int i = 1; i < n + 1; i++) {
    long long int x = query(1, 1, n, arr[i], n);
    ltr[i] = x;
    update(1, 1, n, arr[i]);
    ans += rtl[i] * ltr[i];
  }
  cout << ans << endl;
  return 0;
}

```