Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

int main(){
  int q, query, x, L, R;
  scanf("%d", &q);
  multiset<int> S;
  for(int i = 0; i < q; i++){
    scanf("%d", &query);
    if(query == 3){
      scanf("%d %d", &L,&R);
      auto it = S.lower_bound(L);
      for(auto i = it; i != S.end() && *i <= R; i++){
        printf("%d\n", *i);
      }
    }else{
      scanf("%d", &x);
      if(query == 0){
        S.insert(x);
        printf("%d\n", S.size());
      }else if(query == 1){
        printf("%d\n", S.count(x));
      }else{
        S.erase(x);
      }
    }
  }
  return 0;
}