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 JAVA (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {
	static int cnt=0;
	public static void Merge(int a[],int start,int end){
		if (start>=end) return;
		int center=(start+end)/2;
		Merge(a,start,center);
		Merge(a,center+1,end);
		int n1=center-start+1;
		int n2=end-center;
		int[] L=new int[n1];
		int[] R=new int[n2];
		for (int i=0;i<n1;i++){L[i]=a[start+i];}
		for (int i=0;i<n2;i++){R[i]=a[center+1+i];}
		int i=0; int j=0;
		for (int k=start;k<=end;k++){
			if (i==n1) a[k]=R[j++];
			else if (j==n2) a[k]=L[i++];
			else if (L[i]<=R[j]) a[k]=L[i++];
			else {a[k]=R[j++]; cnt+=n1-i;}
		}
	}
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
	    int n = in.nextInt();
	    int[] A = new int[n];
	    for (int i = 0; i<n; i++) {
	      A[i] = in.nextInt();
	    }
		Merge(A,0,n-1);
		System.out.println(cnt);
	}
}