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<bits/stdc++.h>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define HUGE_NUM 1000000000000000000
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;



map<int,int> MAP;

int main(){

	int num_query;
	scanf("%d",&num_query);

	int command;
	int a,b,sum = 0;

	for(int loop = 0; loop < num_query; loop++){

		scanf("%d",&command);

		switch(command){
		case 0:

			scanf("%d",&a);
			MAP[a]++;
			sum++;
			printf("%d\n",sum);
			break;

		case 1:

			scanf("%d",&a);
			printf("%d\n",MAP[a]);

			break;
		case 2:

			scanf("%d",&a);
			sum -= MAP[a];
			MAP[a] = 0;

			break;
		case 3:

			scanf("%d %d",&a,&b);

			auto at = MAP.lower_bound(a);
			auto right = MAP.upper_bound(b);

			while(at != right){

				for(int i = 0; i < at->second; i++){

					printf("%d\n",at->first);
				}
				at++;
			}
			break;
		}
	}

	return 0;
}

