Problem: 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 $L \leq x \leq R$.

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 $2,000,000$

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.

Example

Input

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


Output

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

Solution (in JAVA):

Solution:
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;
import java.util.SortedMap;
import java.util.SortedSet;
import java.util.TreeMap;
import java.util.TreeSet;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int q = sc.nextInt();
		int count = 0;
		TreeMap<Integer, Integer> map = new TreeMap<>();
		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < q; i++) {
			int op = sc.nextInt();
			int y = sc.nextInt();
			switch (op) {
				case 0:
					if (!map.keySet().contains(y)) {
						map.put(y, 1);
					} else {
						map.put(y, map.get(y) + 1);
					}
					sb.append(++count);
					sb.append('\n');
					break;
				case 1:
					if (map.keySet().contains(y)) {
						sb.append(map.get(y));
					} else {
						sb.append('0');
					}
					sb.append('\n');
					break;
				case 2:
					if (map.keySet().contains(y)) {
						int num2 = map.get(y);
						map.remove(y);
						count -= num2;
					}
					break;
				case 3:
					int z = sc.nextInt() + 1;
					Set<Integer> subset = map.subMap(y, z).keySet();
					for (int num : subset) {
						int end = map.get(num);
						for (int j = 0; j < end; j++) {
							sb.append(num);
							sb.append('\n');
						}
					}
					break;
			}
		}
		System.out.print(sb.toString());
	}
}