#include <bits/stdc++.h>
using namespace std;
int n;
long long bit1[1000005];
long long bit2[1000005];
void update(long long v[], int idx, int val) {
  while (idx <= n) {
    v[idx] += val;
    idx += (idx & -idx);
  }
}
long long query(long long v[], int idx) {
  long long sum = 0;
  while (idx) {
    sum += v[idx];
    idx -= (idx & -idx);
  }
  return sum;
}
int main() {
  scanf("%d", &n);
  set<int> s;
  static int arr[1000005];
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
    s.insert(arr[i]);
  }
  unordered_map<int, int> mask;
  int cnt = 1;
  for (auto each : s) {
    mask[each] = cnt++;
  }
  for (int i = 1; i <= n; i++) {
    arr[i] = mask[arr[i]];
  }
  static long long lss[1000005];
  for (int i = n; i >= 1; i--) {
    lss[arr[i]] = query(bit1, arr[i]);
    update(bit1, arr[i], 1);
  }
  long long ans = 0;
  for (long long i = n; i >= 1; i--) {
    ans += query(bit2, arr[i]);
    update(bit2, arr[i], lss[arr[i]]);
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please formulate 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;
int n;
long long bit1[1000005];
long long bit2[1000005];
void update(long long v[], int idx, int val) {
  while (idx <= n) {
    v[idx] += val;
    idx += (idx & -idx);
  }
}
long long query(long long v[], int idx) {
  long long sum = 0;
  while (idx) {
    sum += v[idx];
    idx -= (idx & -idx);
  }
  return sum;
}
int main() {
  scanf("%d", &n);
  set<int> s;
  static int arr[1000005];
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
    s.insert(arr[i]);
  }
  unordered_map<int, int> mask;
  int cnt = 1;
  for (auto each : s) {
    mask[each] = cnt++;
  }
  for (int i = 1; i <= n; i++) {
    arr[i] = mask[arr[i]];
  }
  static long long lss[1000005];
  for (int i = n; i >= 1; i--) {
    lss[arr[i]] = query(bit1, arr[i]);
    update(bit1, arr[i], 1);
  }
  long long ans = 0;
  for (long long i = n; i >= 1; i--) {
    ans += query(bit2, arr[i]);
    update(bit2, arr[i], lss[arr[i]]);
  }
  printf("%lld\n", ans);
  return 0;
}

```