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: Accepted)

import java.util.*;

class Main{
	static long c=0;
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int A[] = new int[n];
		for(int i=0; i<n; i++){
			A[i] = sc.nextInt();
		}
		mergeSort(A, 0, n);
		System.out.println(c);
	}
	static 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);
		}
	}
	static void merge(int A[], int left, int mid, int right){
		int n1 = mid - left;
		int n2 = right - mid;
		int L[] = new int[n1+1];
		int R[] = new int[n2+1];
		for(int i=0; i<n1; i++){
			L[i] = A[left + i];
		}
		for(int i=0; i<n2; i++){
			R[i] = A[mid + i];
		}
		L[n1] = R[n2] = Integer.MAX_VALUE;
		int i, j;
		i = j = 0;
		for(int k=left; k<right; k++){
			if(L[i] <= R[j]){
				A[k] = L[i++];
			} else{
				A[k] = R[j++];
				c += (n1 - i);
			}
		}
	}
}