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: Time Limit Exceeded)

import java.util.Scanner;

class Main {
	public static void main (String[] args) {
		Scanner scan = new Scanner (System.in);
		final int n = scan.nextInt();
		int[] sequence = new int[n];
		
		for (int i = 0; i < n; i++) {
			sequence[i] = scan.nextInt();
		}
		scan.close();
		
		bubbleSort(sequence);
		
	}
	
	static void bubbleSort (int[] sequence) {
		long cnt = 0;
		int n = sequence.length - 1;
		
		for (int i = 0; i < n; i++) {
			cnt += (long)Inversions(i, sequence);
		}
		
		System.out.println(cnt);
	}
	
	static int Inversions (int i, int[] sequence) {
		int cnt = 0;
		int n = sequence.length;
		
		for (int j = i + 1; j < n; j++) {
			if ( sequence[i] > sequence[j]) {
				cnt++;
			}
		}
		
		return cnt;
	}
}