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<bits/stdc++.h>
using namespace std;
multiset<int> s;
int main()
{
	int n;
	scanf("%d",&n);
	while(n--)
	{
		int x, y;
		scanf("%d%d",&x,&y);
		if(x == 0) 
		{
			s.insert(y);
			printf("%d\n",s.size());
		}
		else if(x == 1) printf("%d\n",s.count(y));
		else if(x == 2) s.erase(y);
		else if(x == 3)
		{
			int l, r;
			l = y;
			scanf("%d",&r);
			multiset<int>::iterator it, it2;
			it = s.lower_bound(l);
			it2 = s.upper_bound(r);
			for(; it!=it2; it++) printf("%d\n",*it);
		}
	}
	return 0;
}