#include <bits/stdc++.h>
using namespace std;
int num[1000010], id[1000010];
long long tree[2][1000010];
int bs(int n, int key) {
  int l = 0, r = n, mid;
  while (l < r) {
    mid = (l + r) / 2;
    if (num[mid] >= key)
      r = mid;
    else
      l = mid + 1;
  }
  return l;
}
void add(long long arr[], int pos, long long n) {
  while (pos < 1000010) {
    arr[pos] += n;
    pos += pos & (-pos);
  }
}
long long query(long long arr[], int pos) {
  long long ans = 0;
  while (pos > 0) {
    ans += arr[pos];
    pos -= pos & (-pos);
  }
  return ans;
}
int main() {
  int n;
  long long ans = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &num[i]), id[i] = num[i];
  sort(num, num + n);
  for (int i = 0, tmp; i < n; ++i)
    tmp = n - bs(n, id[i]), ans += query(tree[1], tmp),
    add(tree[1], tmp, query(tree[0], tmp)), add(tree[0], tmp, 1);
  cout << ans;
  return 0;
}

### 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 num[1000010], id[1000010];
long long tree[2][1000010];
int bs(int n, int key) {
  int l = 0, r = n, mid;
  while (l < r) {
    mid = (l + r) / 2;
    if (num[mid] >= key)
      r = mid;
    else
      l = mid + 1;
  }
  return l;
}
void add(long long arr[], int pos, long long n) {
  while (pos < 1000010) {
    arr[pos] += n;
    pos += pos & (-pos);
  }
}
long long query(long long arr[], int pos) {
  long long ans = 0;
  while (pos > 0) {
    ans += arr[pos];
    pos -= pos & (-pos);
  }
  return ans;
}
int main() {
  int n;
  long long ans = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &num[i]), id[i] = num[i];
  sort(num, num + n);
  for (int i = 0, tmp; i < n; ++i)
    tmp = n - bs(n, id[i]), ans += query(tree[1], tmp),
    add(tree[1], tmp, query(tree[0], tmp)), add(tree[0], tmp, 1);
  cout << ans;
  return 0;
}

```