Problem:
Multi-Set

For a set $S$ of integers, perform a sequence of the following operations. Note that multiple elements can have equivalent values in $S$.

insert($x$): Insert $x$ to $S$ and report the number of elements in $S$ after the operation.

find($x$): Report the number of $x$ in $S$.

delete($x$): Delete all $x$ from $S$.

dump($L$, $R$): Print elements $x$ in $S$ such that $L \leq x \leq R$.

Input

The input is given in the following format.

$q$
$query_1$
$query_2$
:
$query_q$

Each query $query_i$ is given by

0 $x$

or

1 $x$

or

2 $x$

or

3 $L$ $R$

where the first digits 0, 1, 2 and 3 represent insert, find, delete and dump operations respectively.

Output

For each insert operation, print the number of elements in $S$.

For each find operation, print the number of specified elements in $S$.

For each dump operation, print the corresponding elements in ascending order. Print an element in a line.

Constraints

$1 \leq q \leq 200,000$

$0 \leq x \leq 1,000,000,000$

The total number of elements printed by dump operations does not exceed $1,000,000$

The sum of numbers printed by find operations does not exceed $2,000,000$

Sample Input 1

10
0 1
0 1
0 2
0 3
2 2
1 1
1 2
1 3
0 4
3 1 4

Sample Output 1

1
2
3
4
2
0
1
4
1
1
3
4

Solution in JAVA (judge verdict: Wrong Answer)

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.TreeSet;
public class Main {
	public static void main(String[] args){
		int i, j, q, order, x, counter = 0;
		Map<Integer, Integer> map = new HashMap<Integer, Integer>();
		TreeSet<Integer> set = new TreeSet<Integer>();
		try (Scanner sc = new Scanner(System.in)) {
			q = sc.nextInt();
			for(i = 0; i < q; i++) {
				order = sc.nextInt();
				x = sc.nextInt();
				switch(order) {
				case 0 :
					counter++;
					if(!set.add(x)) {
						if(map.containsKey(x)) {
							map.compute(x, (key, old) -> old + 1);
						}else {
							map.put(x, 1);
						}
						
					}
					System.out.println(map.values());
					
					
					//System.out.println(counter);
					break;
				case 1 :
					if(set.contains(x)) {
						if(map.containsKey(x)) {
							System.out.println(map.get(x) + 1);
						}else {
							System.out.println(1);
						}
						
					}else {
						System.out.println(0);
					}
					break;
				case 2 :
					if(set.remove(x)) {
						if(map.containsKey(x)) {
							counter -= map.remove(x);
						}
					}
					break;
				case 3 :
					for(Integer output : set.subSet(x, true, sc.nextInt(), true)) {
						System.out.println(output);
						if(map.containsKey(output)) {
							for(j = 0; j < map.get(output); j++) {
								System.out.println(output);
							}
						}
					}
					break;
				}
			}
		}
	}
}