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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

public class Main {
	
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		
		final int n = sc.nextInt();
		
		TreeSet<Integer> set = new TreeSet<Integer>();
		int count = 0;
		for(int i = 0; i < n; i++){
			final int a = sc.nextInt();
			
			count += i - set.headSet(a).size();
			set.add(a);
		}
		
		System.out.println(count);
		
	}
	
	public static class Scanner {
	    private BufferedReader br;
	    private StringTokenizer tok;

	    public Scanner(InputStream is) throws IOException {
	        br = new BufferedReader(new InputStreamReader(is));
	    }

	    private void getLine() throws IOException {
	        while (!hasNext()) { tok = new StringTokenizer(br.readLine()); }
	    }

	    private boolean hasNext() {
	        return tok != null && tok.hasMoreTokens();
	    }

	    public String next() throws IOException {
	        getLine(); return tok.nextToken();
	    }

	    public int nextInt() throws IOException {
	        return Integer.parseInt(next());
	    }
	    
	    public long nextLong() throws IOException {
	    	return Long.parseLong(next());
	    }
	    
	    public void close() throws IOException {
	        br.close();
	    }
	}

}