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

    private static long merge(int[] a1, int[] a2, int a[]) {
        long cnt = 0;
        int i1 = 0;
        int i2 = 0;
        while (i1 < a1.length || i2 < a2.length) {
            if (i1 < a1.length && (i2 >= a2.length || a1[i1] <= a2[i2])) {
                a[i1 + i2] = a1[i1++];
            } else {
                a[i1 + i2] = a2[i2++];
                cnt += a1.length - i1;
            }
        }
        return cnt;
    }

    private static long mergeSort(int[] a) {
        if (a.length < 2) {
            return 0;
        }
        long cnt = 0;
        int n1 = a.length / 2;
        int n2 = a.length - n1;
        int[] a1 = new int[n1];
        int[] a2 = new int[n2];
        for (int i = 0; i < n1; i++) {
            a1[i] = a[i];
        }
        for (int i = 0; i < n2; i++) {
            a2[i] = a[n1 + i];
        }
        cnt += mergeSort(a1);
        cnt += mergeSort(a2);
        return cnt + merge(a1, a2, a);
    }
}