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>

using namespace std;

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