#include <bits/stdc++.h>
using namespace std;
int tree[4 * 1000007], arr[1000007], arr2[1000007], chhoto[1000007],
    boro[1000007];
unordered_map<int, int> mp;
void update(int node, int l, int r, int i) {
  if (l > i or r < i) return;
  if (l == i and r == i) {
    tree[node]++;
    return;
  }
  update(2 * node, l, (l + r) / 2, i);
  update(2 * node + 1, (l + r) / 2 + 1, r, i);
  tree[node] = tree[2 * node] + tree[2 * node + 1];
}
int query(int node, int l, int r, int i, int j) {
  if (l > j or r < i) return 0;
  if (l >= i and r <= j) return tree[node];
  int q1 = query(2 * node, l, (l + r) / 2, i, j);
  int q2 = query(2 * node + 1, (l + r) / 2 + 1, r, i, j);
  return q1 + q2;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]), arr2[i] = arr[i];
  sort(arr2, arr2 + n);
  int cnt = 1;
  for (int i = 0; i < n; i++)
    if (!mp[arr2[i]]) mp[arr2[i]] = cnt++;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long bigger_before = query(1, 1, n, mp[arr[i]] + 1, cnt - 1);
    long long smaller_after = (mp[arr[i]] - 1) - (i - bigger_before);
    ans += bigger_before * smaller_after;
    update(1, 1, n, mp[arr[i]]);
  }
  printf("%I64d\n", 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;
int tree[4 * 1000007], arr[1000007], arr2[1000007], chhoto[1000007],
    boro[1000007];
unordered_map<int, int> mp;
void update(int node, int l, int r, int i) {
  if (l > i or r < i) return;
  if (l == i and r == i) {
    tree[node]++;
    return;
  }
  update(2 * node, l, (l + r) / 2, i);
  update(2 * node + 1, (l + r) / 2 + 1, r, i);
  tree[node] = tree[2 * node] + tree[2 * node + 1];
}
int query(int node, int l, int r, int i, int j) {
  if (l > j or r < i) return 0;
  if (l >= i and r <= j) return tree[node];
  int q1 = query(2 * node, l, (l + r) / 2, i, j);
  int q2 = query(2 * node + 1, (l + r) / 2 + 1, r, i, j);
  return q1 + q2;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]), arr2[i] = arr[i];
  sort(arr2, arr2 + n);
  int cnt = 1;
  for (int i = 0; i < n; i++)
    if (!mp[arr2[i]]) mp[arr2[i]] = cnt++;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long bigger_before = query(1, 1, n, mp[arr[i]] + 1, cnt - 1);
    long long smaller_after = (mp[arr[i]] - 1) - (i - bigger_before);
    ans += bigger_before * smaller_after;
    update(1, 1, n, mp[arr[i]]);
  }
  printf("%I64d\n", ans);
}

```