#include <bits/stdc++.h>
using namespace std;
struct FenwickTree {
  int n;
  long long int tree[1000005];
  void init(int len) {
    n = len;
    for (int i = 1; i <= n; i++) {
      tree[i] = 0;
    }
  }
  void add(int index, long long int value) {
    while (index <= n) {
      tree[index] += value;
      index += (index & -index);
    }
  }
  long long int getPrefix(int index) {
    long long int ans = 0;
    while (index > 0) {
      ans += tree[index];
      index -= (index & -index);
    }
    return ans;
  }
};
FenwickTree BIT1;
FenwickTree BIT2;
int arr[1000005];
map<int, int> compressData;
int main() {
  int n, i, j, count;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
    compressData[arr[i]];
  }
  count = 1;
  for (auto &s : compressData) {
    compressData[s.first] = count++;
  }
  for (i = 1; i <= n; i++) {
    arr[i] = compressData[arr[i]];
  }
  BIT1.init(n);
  BIT2.init(n);
  long long int ans = 0;
  for (i = n; i >= 1; i--) {
    ans += BIT2.getPrefix(arr[i]);
    BIT2.add(arr[i], BIT1.getPrefix(arr[i]));
    BIT1.add(arr[i], 1);
  }
  cout << 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;
struct FenwickTree {
  int n;
  long long int tree[1000005];
  void init(int len) {
    n = len;
    for (int i = 1; i <= n; i++) {
      tree[i] = 0;
    }
  }
  void add(int index, long long int value) {
    while (index <= n) {
      tree[index] += value;
      index += (index & -index);
    }
  }
  long long int getPrefix(int index) {
    long long int ans = 0;
    while (index > 0) {
      ans += tree[index];
      index -= (index & -index);
    }
    return ans;
  }
};
FenwickTree BIT1;
FenwickTree BIT2;
int arr[1000005];
map<int, int> compressData;
int main() {
  int n, i, j, count;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
    compressData[arr[i]];
  }
  count = 1;
  for (auto &s : compressData) {
    compressData[s.first] = count++;
  }
  for (i = 1; i <= n; i++) {
    arr[i] = compressData[arr[i]];
  }
  BIT1.init(n);
  BIT2.init(n);
  long long int ans = 0;
  for (i = n; i >= 1; i--) {
    ans += BIT2.getPrefix(arr[i]);
    BIT2.add(arr[i], BIT1.getPrefix(arr[i]));
    BIT1.add(arr[i], 1);
  }
  cout << ans;
  return 0;
}

```