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
#include<iostream>
#include<set>
#include<algorithm>
using namespace std;
int main()
{
	int n;
	cin >> n;
	multiset<int> p;
	while (n--)
	{
		int oper;
		cin >> oper;
		switch (oper)
		{
			int num;
		case 0:
			cin >> num;
			p.insert(num);
			cout << p.size() << endl;
			break;
		case 1:
			cin >> num;
			cout << p.count(num) << endl;
			break;
		case 2:
			cin >> num;
			p.erase(num);
			break;
		case 3:
			int head, end;
			cin >> head >> end;
			set<int>::iterator p1 = p.lower_bound(head);
			set<int>::iterator p2 = p.upper_bound(end);
			for (;p1 != p2;p1++)
			{
				cout << *p1 << endl;
			}

			
		}
	}
}
