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;
typedef long long ll;

// #define DEBUG

int main()
{

#ifdef DEBUG
    cout << "DEBUG MODE" << endl;
    ifstream in("input.txt"); //for debug
    cin.rdbuf(in.rdbuf());    //for debug
#endif

    int n, k, x, y;
    cin >> n;
    // set<int> workspace;
    multiset<int> workspace;

    for (int i = 0; i < n; i++)
    {
        cin >> k;
        if (k == 0)
        {
            cin >> x;
            workspace.insert(x);
            cout << workspace.size() << "\n";
        }
        else if (k == 1)
        {
            cin >> x;
            cout << workspace.count(x) << "\n";
            // auto z = workspace.find(x);
            // cout << ((z == workspace.end()) ? 0 : 1) << "\n";
        }
        else if (k == 2)
        {
            cin >> x;
            workspace.erase(x);
        }
        else if (k == 3)
        {
            cin >> x >> y;
            for (auto itr = workspace.lower_bound(x); itr != workspace.upper_bound(y); itr++)
                cout << *itr << "\n";
        }
    }
    return 0;
}

// getline(cin,s);
// a = strtok(s, ",");  ","で分解
// for (int i = 0; i < n; i++)
