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 counter = 0;
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int [] ls = new int[n];
        for(int x = 0; x < n; x++){
            ls[x] = scan.nextInt();
        }
        merge_sort(ls, 0, n);
        System.out.println(counter);
    }
    public static void merge_sort (int[] lst, int left, int right) {
        if(left + 1 < right){
            int mid = (left + right)/2;
            merge_sort(lst, left, mid);
            merge_sort(lst, mid, right);
            merge(lst, left, mid, right);
        }
    }
    public 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 l = 0;
        int r = 0;
        for(int i = left; i < right; i++){
            if (L[l] <= R[r]){
                A[i] = L[l++];
                counter = counter + r;
            }
            else{
                A[i] = R[r++];
            }
        }
    }
}