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 ll;
int L[MAX / 2 + 2],R[MAX / 2 + 2];
ll merge(int A[], int n,int left,int mid, int right){
	int i,j,k;
	ll cnt = 0;
	int n1 = mid - left;
	int n2 = right-mid;
	for(i = 0; i < n1;i++){
		L[i] = A[left + i];
	}
	for(i = 0; i < n2; i++){
		R[i] = A[mid + i];
	}
	L[n1] = R[n2] = SENTINEL;
	i = 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;
}
ll mergeSort(int A[], int n,int left, int right){
	int mid;
	ll 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 (void){
	
	int A[MAX],n,i;
	scanf("%d",&n);
	for(i = 0;i < n;i++){
		scanf("%d",&A[i]);
	}
	ll ans = mergeSort(A,n,0,n);
	printf("%ld\n",ans);
	return 0;
}