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;

int main(){
	int n;
	scanf("%d", &n);
	multiset<int> s;
	for(int i = 0; i < n; i++) {
		int id, x;
		scanf("%d%d", &id, &x);
		if(id == 0) {
			s.insert(x);
			printf("%lu\n", s.size());
		}else if(id == 1) {
			printf("%lu\n", s.count(x));
		}else if(id == 2) {
			s.erase(x);
		}else if(id == 3) {
			int y;
			scanf("%d", &y);
			auto itr = s.lower_bound(x);
			for(; itr != s.end() && *itr <= y; itr++)
				printf("%d\n", *itr);
		}
	}
}