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 <algorithm>
#include<iostream>
#include<vector>
#include<deque>
#include<queue>
#include<list>
#include<stack>
#include<map>
#include<set>
#include<string>
#include <sstream>

#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<limits.h>

using namespace std;


int main(){
  long ii,jj,kk;
  vector<int> a;
  multiset<int> s;
  int q,c,x,l,r;

  cin >> q;
  multiset<int>::iterator it;
  multiset<int>::iterator lit,rit;  
  for(ii=0;ii<q;ii++){
    cin >> c >> x;
    if(c == 0){
      s.insert(x);
      cout << s.size() << endl;
    }
    else if(c == 1){
      it = s.find(x);
      if(it != s.end()){
        cout << s.count(x) << endl;
      }
      else{
        cout << 0 << endl;
      }
    }
    else if(c == 2){
      s.erase(x);
    }
    else{
      l = x;
      cin >> r;
      lit = s.lower_bound(l);
      rit = s.lower_bound(r);

      while(lit != rit){
        cout << *lit << endl;
        lit++;
      }
      int cnt=0;
      if(*lit <= r && lit!=s.end()){
        for(jj=0;jj<s.count(r);jj++){
          cout << *lit << endl;
          lit++;
        }

      }
    }
  }
  
  return 0;
}

