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 <iterator>
#include <vector>
#include <algorithm>
#include <string>
#include <functional>
#include <set>

using namespace std;

int main()
{
    int q;
    cin >> q;

    multiset<int> S;
    int cmd, x, L, R;
    while (q--)
    {
        cin >> cmd;
        if (cmd == 0)
        {
            cin >> x;
            S.insert(x);
            cout << S.size() << endl;
        }
        else if (cmd == 1)
        {
            cin >> x;
            // cout << (S.find(x) != S.end()) << endl;
            cout << S.count(x) << endl;
        }
        else if (cmd == 2)
        {
            cin >> x;
            S.erase(x);
            // remove(S.begin(), S.end(), x);
        }
        else
        {
            cin >> L >> R;
            multiset<int>::iterator it1 = S.lower_bound(L), it2 = S.upper_bound(R);
            while (it1 != it2)
            {
                cout << *it1 << endl;
                it1++;
            }

        }
        
        
    }
}
