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: Wrong Answer)

#include<stdio.h>
#include<limits.h>

int compTimes = 0;

void merge(int A[], int left, int mid, int right){
	int n1 = mid - left;
	int n2 = right - mid;
	int L[n1+1], R[n2+1];
	int i, j, k;
	for( i = 0; i < n1; i++ ){ L[i] = A[left+i]; }
	for( i = 0; i < n2; i++ ){ R[i] = A[mid+i]; }
	L[n1] = INT_MAX;
	R[n2] = INT_MAX;
	i = 0;
	j = 0;
	int c = 0;
	for( k = left; k < right; k++ ){
		if( L[i] <= R[j] ){
			A[k] = L[i++];
			c++;
		}else{
			if( c < n1 ){ compTimes += n1 - c; }
			A[k] = R[j++];
		}
	}
	return;
}

void mergeSort(int A[], int left, int right){
	if( left+1 <  right ){
		int mid = (left + right) / 2;
		mergeSort(A, left, mid);
		mergeSort(A, mid, right);
		merge(A, left, mid, right);
	}
	return;
}

int main(){
	int A[200000], S[200000];
	int n, i, j, count = 0;
	scanf("%d", &n);
	for( i = 0; i < n; i++ ){
		scanf("%d", &A[i]);
		S[i] = A[i];
	}
	mergeSort(S, 0, n);
	printf("%d\n", compTimes);
	return 0;
}