Problem:
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The Number of Inversions

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.

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.

output

Print the number of inversions in a line.

Constraints

$ 1 \leq n \leq 200,000$

$ 0 \leq a_i \leq 10^9$

$a_i$ are all different

Sample Input 1

5
3 5 2 1 4

Sample Output 1

6

Sample Input 2

3
3 1 2

Sample Output 2

2

Solution in C (judge verdict: Accepted)

#include <stdio.h>
#define MAX 200000
#define SENTINEL 2000000000
typedef long long llong;
int L[MAX/2+2],R[MAX/2+2];

llong merge(int A[],int n, int left, int mid, int right)
{
	int i,j,k;
	llong cnt=0;
	int n1 = mid -left;
	int n2 = right - mid;
	for(i=0; i<n1;i++)
	{
		L[i] = A[left+i];
	}
	for(int i=0;i<n2;i++)
	{
		R[i] = A[mid+i];
	}
	L[n1]=R[n2]=SENTINEL;
	i=0;
	j=0;
	for(k=left;k<right;k++)
	{
		if(L[i]<=R[j])
		{
			A[k]=L[i++];
		}
		else
		{
			A[k]=R[j++];
			cnt += n1-i;
		}
	}
	return cnt;

}

llong mergeSort(int A[],int n,int left,int right)
{
	int mid;
	llong v1, v2, v3;
	if(left+1<right)
	{
		mid = (left+right)/2;
		v1=mergeSort(A,n,left,mid);
		v2=mergeSort(A,n,mid,right);
		v3=merge(A,n,left,mid,right);
		return v1+v2+v3;
	}else return 0;
}

int main()
{
	int A[MAX],n,i;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
		scanf("%d",&A[i]);
	}
	llong ans=mergeSort(A,n,0,n);

	printf("%lld\n",ans);
	return 0;
}