Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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.Scanner;



public class Main {
	static long count;

	public static void merge(int[] arr, int right, int mid, int left) {
		int n1 = mid - left, n2 = right - mid;
		int[] l = new int[n1 + 1], r = new int[n2 + 1];
		for (int i = 0; i < n1; i++) {
			l[i] = arr[left + i];
		}
		for (int i = 0; i < n2; i++) {
			r[i] = arr[mid + i];
		}
		l[n1] = Integer.MAX_VALUE;
		r[n2] = Integer.MAX_VALUE;
		int i = 0, j = 0;
		for (int k = left; k < right; k++) {
			if (l[i] <= r[j]) {
				arr[k] = l[i++];
			} else {
				arr[k] = r[j++];
				count += n1 - i;
			}
		}
	}
	public static void mergeSort(int[] arr, int left, int right) {
		if (left + 1 < right) {
			int mid = (left + right) / 2;
			mergeSort(arr, left, mid);
			mergeSort(arr, mid, right);
			merge(arr, right, mid, left);
		}
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] arr = new int[n];
		for (int i = 0; i < n; i++) {
			arr[i] = sc.nextInt();
		}
		mergeSort(arr, 0, n);
		System.out.println(count);
	}
}