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(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;
}

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