#include <bits/stdc++.h>
using namespace std;
long long int dp1[1000010], dp2[1000010], a[1000010], b[1000010],
    ltree[1000010], gtree[1000010];
void update(long long int *tree, long long int n, long long int index,
            long long int val) {
  while (index <= n) {
    tree[index] += val;
    index += (index & (-index));
  }
}
long long int query(long long int *tree, long long int index) {
  long long int sum = 0;
  while (index > 0) {
    sum += tree[index];
    index -= (index & (-index));
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  long long int i;
  for (i = 1; i <= n; i++) {
    ltree[i] = 0;
    gtree[i] = 0;
  }
  for (i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(a, a + n);
  for (i = 0; i < n; i++) {
    long long int *p = lower_bound(a, a + n, b[i]);
    b[i] = (p - a);
    b[i]++;
  }
  for (i = 0; i < n; i++) {
    dp1[i + 1] = query(gtree, n) - query(gtree, b[i]);
    update(gtree, n, b[i], +1ll);
  }
  for (i = (n - 1); i >= 0; i--) {
    dp2[i + 1] = query(ltree, b[i] - 1);
    update(ltree, n, b[i], +1ll);
  }
  long long int ans = 0;
  for (i = 1; i <= n; i++) {
    ans += dp1[i] * dp2[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;
long long int dp1[1000010], dp2[1000010], a[1000010], b[1000010],
    ltree[1000010], gtree[1000010];
void update(long long int *tree, long long int n, long long int index,
            long long int val) {
  while (index <= n) {
    tree[index] += val;
    index += (index & (-index));
  }
}
long long int query(long long int *tree, long long int index) {
  long long int sum = 0;
  while (index > 0) {
    sum += tree[index];
    index -= (index & (-index));
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  long long int i;
  for (i = 1; i <= n; i++) {
    ltree[i] = 0;
    gtree[i] = 0;
  }
  for (i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(a, a + n);
  for (i = 0; i < n; i++) {
    long long int *p = lower_bound(a, a + n, b[i]);
    b[i] = (p - a);
    b[i]++;
  }
  for (i = 0; i < n; i++) {
    dp1[i + 1] = query(gtree, n) - query(gtree, b[i]);
    update(gtree, n, b[i], +1ll);
  }
  for (i = (n - 1); i >= 0; i--) {
    dp2[i + 1] = query(ltree, b[i] - 1);
    update(ltree, n, b[i], +1ll);
  }
  long long int ans = 0;
  for (i = 1; i <= n; i++) {
    ans += dp1[i] * dp2[i];
  }
  cout << ans << endl;
  return 0;
}

```