#include <bits/stdc++.h>
using namespace std;
int a[1000050], temp_a[1000050], gt[1000050], le[1000050], seg_tree[4000050];
map<int, int> to_new_num;
int query(int q_l, int q_r, int l, int r, int id) {
  if (q_l <= l && r <= q_r) return seg_tree[id];
  if (q_r < l || r < q_l) return 0;
  int mid = (l + r) / 2;
  return query(q_l, q_r, l, mid, 2 * id) +
         query(q_l, q_r, mid + 1, r, 2 * id + 1);
}
void update(int target, int l, int r, int id) {
  seg_tree[id]++;
  if (l == r) return;
  int mid = (l + r) / 2;
  if (target <= mid)
    update(target, l, mid, 2 * id);
  else
    update(target, mid + 1, r, 2 * id + 1);
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    temp_a[i] = a[i];
  }
  sort(temp_a, temp_a + n);
  int cnt = 1, temp = temp_a[0];
  to_new_num[temp_a[0]] = cnt;
  for (int i = 1; i < n; i++) {
    if (temp != temp_a[i]) {
      to_new_num[temp_a[i]] = ++cnt;
    }
  }
  for (int i = 0; i < n; i++) a[i] = to_new_num[a[i]];
  memset(seg_tree, 0, sizeof(seg_tree));
  for (int i = 0; i < n; i++) {
    gt[i] = query(a[i] + 1, cnt, 1, cnt, 1);
    update(a[i], 1, cnt, 1);
  }
  memset(seg_tree, 0, sizeof(seg_tree));
  for (int i = n - 1; i >= 0; i--) {
    le[i] = query(1, a[i] - 1, 1, cnt, 1);
    update(a[i], 1, cnt, 1);
  }
  long long int ans = 0;
  for (int i = 0; i < n; i++) ans += (long long int)gt[i] * le[i];
  cout << ans << '\n';
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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[1000050], temp_a[1000050], gt[1000050], le[1000050], seg_tree[4000050];
map<int, int> to_new_num;
int query(int q_l, int q_r, int l, int r, int id) {
  if (q_l <= l && r <= q_r) return seg_tree[id];
  if (q_r < l || r < q_l) return 0;
  int mid = (l + r) / 2;
  return query(q_l, q_r, l, mid, 2 * id) +
         query(q_l, q_r, mid + 1, r, 2 * id + 1);
}
void update(int target, int l, int r, int id) {
  seg_tree[id]++;
  if (l == r) return;
  int mid = (l + r) / 2;
  if (target <= mid)
    update(target, l, mid, 2 * id);
  else
    update(target, mid + 1, r, 2 * id + 1);
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    temp_a[i] = a[i];
  }
  sort(temp_a, temp_a + n);
  int cnt = 1, temp = temp_a[0];
  to_new_num[temp_a[0]] = cnt;
  for (int i = 1; i < n; i++) {
    if (temp != temp_a[i]) {
      to_new_num[temp_a[i]] = ++cnt;
    }
  }
  for (int i = 0; i < n; i++) a[i] = to_new_num[a[i]];
  memset(seg_tree, 0, sizeof(seg_tree));
  for (int i = 0; i < n; i++) {
    gt[i] = query(a[i] + 1, cnt, 1, cnt, 1);
    update(a[i], 1, cnt, 1);
  }
  memset(seg_tree, 0, sizeof(seg_tree));
  for (int i = n - 1; i >= 0; i--) {
    le[i] = query(1, a[i] - 1, 1, cnt, 1);
    update(a[i], 1, cnt, 1);
  }
  long long int ans = 0;
  for (int i = 0; i < n; i++) ans += (long long int)gt[i] * le[i];
  cout << ans << '\n';
  return 0;
}

```