#include <iostream>
#include <algorithm>
using namespace std;

int arr[200005];
int n;

long long dnc(int from, int to){
    if(from>=to-1)
        return 0;
    
    long long num=0;
    int mid=(from+to)/2;
    num+=dnc(from, mid);
    num+=dnc(mid, to);
    
    int i=from, j=mid, cnt=0;
    while(i<mid && j<to){
        if(arr[i]>arr[j]){
            num+=j-i-cnt++;
            j++;
        }
        else{
            i++;
        }
    }
    
    sort(arr+from, arr+to);
    return num;
}

int main() {
    ios_base::sync_with_stdio(0);
    
    cin >> n;
    for(int i=0; i<n; i++){
        cin >> arr[i];
    }
    
    cout << dnc(0, n) << endl;
    
    return 0;
}
### Prompt

Your task is to create a cpp solution to the following problem:

For a given sequence $A = \\{a_0, a_1, ... a_{n-1}\\}$, the number of pairs $(i, j)$ where $a_i > a_j$ and $i < j$, is called the number of inversions. The number of inversions is equal to the number of swaps of Bubble Sort defined in the following program:


bubbleSort(A)
cnt = 0 // the number of inversions
for i = 0 to A.length-1
for j = A.length-1 downto i+1
if A[j] < A[j-1]
	swap(A[j], A[j-1])
	cnt++

return cnt


For the given sequence $A$, print the number of inversions of $A$. Note that you should not use the above program, which brings Time Limit Exceeded.

Constraints

* $ 1 \leq n \leq 200,000$
* $ 0 \leq a_i \leq 10^9$
* $a_i$ are all different

Input

In the first line, an integer $n$, the number of elements in $A$, is given. In the second line, the elements $a_i$ ($i = 0, 1, .. n-1$) are given separated by space characters.

Examples

Input

5
3 5 2 1 4


Output

6


Input

3
3 1 2


Output

2

### Response

```cpp
#include <iostream>
#include <algorithm>
using namespace std;

int arr[200005];
int n;

long long dnc(int from, int to){
    if(from>=to-1)
        return 0;
    
    long long num=0;
    int mid=(from+to)/2;
    num+=dnc(from, mid);
    num+=dnc(mid, to);
    
    int i=from, j=mid, cnt=0;
    while(i<mid && j<to){
        if(arr[i]>arr[j]){
            num+=j-i-cnt++;
            j++;
        }
        else{
            i++;
        }
    }
    
    sort(arr+from, arr+to);
    return num;
}

int main() {
    ios_base::sync_with_stdio(0);
    
    cin >> n;
    for(int i=0; i<n; i++){
        cin >> arr[i];
    }
    
    cout << dnc(0, n) << endl;
    
    return 0;
}
```