#include<bits/stdc++.h>
using namespace std;
int a[200005],L[100001],R[100001];long long cnt=0;
void merge(int l,int mid,int r){
	int n1=mid-l,n2=r-mid;
	for(int i=1;i<=n1;i++) L[i]=a[l+i-1];
	for(int i=1;i<=n2;i++) R[i]=a[mid+i-1];
	int i=1,j=1;L[n1+1]=R[n2+1]=1<<30;
	for(int k=l;k<r;k++){
		if (L[i]<R[j]){
			a[k]=L[i++];
		}
		else{
			a[k]=R[j++];
			cnt+=n1-i+1;
		}
	}
}
void mergesort(int l,int r){
	if (l+1<r){
		int mid =(l+r)/2;
		mergesort(l,mid);
		mergesort(mid,r);
		merge(l,mid,r);
	}
}

int main(){
	int n;cin>>n;
	for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	mergesort(1,n+1);
	printf("%lld\n",cnt);
}
### Prompt

Create a solution in cpp for 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<bits/stdc++.h>
using namespace std;
int a[200005],L[100001],R[100001];long long cnt=0;
void merge(int l,int mid,int r){
	int n1=mid-l,n2=r-mid;
	for(int i=1;i<=n1;i++) L[i]=a[l+i-1];
	for(int i=1;i<=n2;i++) R[i]=a[mid+i-1];
	int i=1,j=1;L[n1+1]=R[n2+1]=1<<30;
	for(int k=l;k<r;k++){
		if (L[i]<R[j]){
			a[k]=L[i++];
		}
		else{
			a[k]=R[j++];
			cnt+=n1-i+1;
		}
	}
}
void mergesort(int l,int r){
	if (l+1<r){
		int mid =(l+r)/2;
		mergesort(l,mid);
		mergesort(mid,r);
		merge(l,mid,r);
	}
}

int main(){
	int n;cin>>n;
	for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	mergesort(1,n+1);
	printf("%lld\n",cnt);
}
```