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 <fstream>
#include <algorithm>
#include <set>
//#define LOCAL
using namespace std;
typedef long long ll;
multiset<ll>s;

int main() {
#ifdef LOCAL
ifstream fin("1.txt");
streambuf*p;
p=cin.rdbuf(fin.rdbuf());
#endif // LOCAL
ll q;
cin>>q;
for(ll i=0;i<q;i++){
    ll a,b;
    cin>>a>>b;
    if(0==a){
        s.insert(b);
        cout<<s.size()<<endl;
    }else if(1==a){
        cout<<s.count(b)<<endl;
    }else if(2==a){
        s.erase(b);
    }else{
        ll t;
        cin>>t;
        set<ll>::iterator i1=s.lower_bound(b);
        set<ll>::iterator i2=s.upper_bound(t);
        for(;i1!=i2;i1++) cout<<(*i1)<<endl;
    }
}

#ifdef LOCAL
fin.close();
#endif // LOCAL
}

