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 CPP):

Solution:
#include<set>
#include<cstdio>
#include<algorithm>
using namespace std;

int main()
{
	multiset<int> cpp; 
	int q, com, o, r;
	scanf("%d", &q);

	while (q--)
	{

		scanf("%d%d", &com, &o);
		switch (com)
		{
		case 0:
			cpp.insert(o);
			printf("%d\n", cpp.size());
			break;

		case 1:
			printf("%d\n", cpp.count(o));
			break;

		case 2:
			cpp.erase(o);
			break;
		case 3:
			scanf("%d", &r);
			set<int>::iterator prev = cpp.lower_bound(o);
			set<int>::iterator next = cpp.upper_bound(r);
			while (prev != next)    
			{
				printf("%d\n", *prev);
				prev++;
			}
			break;
		}
	}
	return 0;
}