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.Scanner;

public class Main {
	static long count = 0;

	public static long 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] = 1000000001;
		R[n2] = 1000000001;
		int i = 0;
		int j = 0;
		for (int k = left; k < right; k++) {
			if (L[i] <= R[j]) {
				A[k] = L[i];
				i = i + 1;
			} else {
				A[k] = R[j];
				j = j + 1;
				count += n1 - i;
			}

		}
		return count;

	}

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

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int[] numbers = new int[n];
		for (int i = 0; i < n; i++) {
			numbers[i] = scanner.nextInt();
		}
		count = mergeSort(numbers, 0, n);
		System.out.println(count);
		scanner.close();
	}

}