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)

/*
 *	argorithms and data structures
 *	lectuire 5 
 *	The Number of Inversions
 */
#include <stdio.h>
#include <limits.h>

#define INFTY INT_MAX
#define N 200000

long long int count = 0;

void merge(int A[], int left, int mid, int right){
	int i, j, k;
	int L[N], R[N];
	int L_elements = mid - left;
	int R_elements = right - mid;

	for(i=0; i<L_elements; i++)
		L[i] = A[left + i];
	L[i] = INFTY;

	for(i=0; i<R_elements; i++)
		R[i] = A[mid + i];
	R[i] = INFTY;


	i = j = k = 0;
	for(k=left; k<right; k++){
		if( L[i] <= R[j] )
			A[k] = L[i++];
		else{
			A[k] = R[j++];
			if(L[i] < INFTY) count += L_elements - i;
		}
	}
}	

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

int main(void){
	int A[N], n, i;

	scanf("%d", &n);
	for(i=0; i<n; i++)
		scanf("%d", &A[i]);

	mergeSort(A, 0, n);

	printf("%llu\n", count);

	return 0;
}