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>

long cnt=0;

void merge(int s[],int n,int left,int mid,int right){
	int i,j,k;
	long L[n];
	long R[n];
	int n1 = mid - left;
	int n2 = right - mid;
	
	for(i = 0;i<n1;i++){
		L[i] = s[left + i];
	}
	for(i = 0;i<n2;i++){
		R[i] = s[mid+i];
	}
	L[n1]=2000000000;
	R[n2]=2000000000;
	i = 0;
	j = 0;
	for(k = left;k<right;k++){
		if(L[i]<R[j]){
			s[k] = L[i++];
		}
		else{ 
			cnt=cnt+(n1-i);
			s[k] = R[j++];
		}
	}
}
	
void mergeSort(int s[],int n,int left,int right){
	if(left+1< right){
		int mid; 
		mid= (left + right)/2;
		mergeSort(s,n,left, mid);
		mergeSort(s,n,mid, right);
		merge(s,n,left,mid, right);
	}
}
	
int main(void){
	int i,n;
	int s[500000];
	
	scanf("%d",&n);
	for(i=0;i<n;i++){
		scanf("%d",&s[i]);
	}
	
	mergeSort(s,n,0,n);
	
	
//	for(i=0;i<n;i++){
//		if(i==n-1){
//			printf("%d\n",s[i]);
//		}
//		else{
//			printf("%d ",s[i]);
//		}
//	}
	
	printf("%ld\n",cnt);
	
	return 0;
}