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.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int[] a = new int[N];
		ArrayList<Integer> list = new ArrayList<Integer>();
		for (int i = 0; i < N; i++) {
			a[i] = sc.nextInt();
			list.add(a[i]);
		}
		Collections.sort(list);
		sc.close();
		BinaryIndexedTree tree = new BinaryIndexedTree(N);

		long ans = 0;
		for (int i = 0; i < N; i++) {
			a[i] = Collections.binarySearch(list, a[i]) + 1;
			ans += i - tree.sum(a[i]);
			tree.add(a[i], 1);
		}
		System.out.println(ans);

	}

	static class BinaryIndexedTree {
		int N;
		int[] bit;

		public BinaryIndexedTree(int N) {
			this.N = N;
			bit = new int[N + 1];

		}

		public int sum(int i) {
			int sum = 0;
			while (i > 0) {
				sum += bit[i];
				i -= i & -i;
			}
			return sum;
		}

		public void add(int i, int x) {
			while (i <= N) {
				bit[i] += x;
				i += i & -i;
			}
		}
	}

}