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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class Main {
    final static int SENTINEL = Integer.MAX_VALUE;

    static long merge(int[] A, int left, int mid, int right) {
        long cnt = 0;
        int n1 = mid - left;
        int n2 = right - mid;

        int[] L = new int[n1 + 1];
        int[] R = new int[n2 + 1];
        System.arraycopy(A, left, L, 0, n1);
        System.arraycopy(A, mid, R, 0, n2);
        L[n1] = SENTINEL;
        R[n2] = SENTINEL;

        int i = 0, j = 0;
        for (int k = left; k < right; k++) {
            if (L[i] <= R[j]) {
                A[k] = L[i];
                i++;
            } else {
                A[k] = R[j];
                j++;
                cnt += n1 - i;
            }
        }
        return cnt;
    }

    static long mergeSort(int[] A, int left, int right) {
        if (right - left > 1) {
            int mid = (left + right) / 2;
            long a = mergeSort(A, left, mid);
            long b = mergeSort(A, mid, right);
            long c = merge(A, left, mid, right);
            return a + b + c;
        } else {
            return 0;
        }
    }

    public static void main(String[] args) {
        try (BufferedReader in = new BufferedReader(new InputStreamReader(System.in))) {
            final int n = Integer.parseInt(in.readLine());
            final int[] A = Arrays.stream(in.readLine().split(" "))
                    .mapToInt(Integer::parseInt).toArray();

            long ret = mergeSort(A, 0, n);

            System.out.println(ret);

        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}