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.Scanner;
import java.util.SortedMap;
import java.util.TreeMap;

class Main {

	public static void main(String[] args) {
		try (Scanner sc = new Scanner(System.in)){
			int times = sc.nextInt();
			TreeMap<Long, Long> map = new TreeMap<>();
			long size = 0l;
			for (int i = 0; i<times; i++){
				int query = sc.nextInt();  //0insert 1find 2delete 3dump
				long x = sc.nextLong();
				switch (query) {
				//insert(x)
				case 0:
					if (!map.containsKey(x)) map.put(x,1l);
					else map.put(x,map.get(x)+1);
					size++;  //毎回size+1
					System.out.println(size);
					break;
				//find(x)
				case 1:
					if (map.get(x)!=null) {
						System.out.println(map.get(x));
					} else System.out.println(0);
					break;
				//delete(x)
				case 2:
					if (map.get(x) != null) {
						size -= map.get(x);  //remove時sizeを減る
						map.remove(x);
					}
					break;
				//dump(x,large)
				case 3:
					long large = sc.nextLong();
					SortedMap<Long,Long> submap = map.subMap(x,true,large,true);
					for (Long l : submap.keySet()) {
						for (int j=0; j<map.get(l); j++) System.out.println(l);   //回数があるだけ出力
					}
					break;
				}
			}
		}
	}
}