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;
using UL = unsigned int;
using ULL = unsigned long long;
using LL = long long;
#define rep(i, n) for(UL i = 0; i < (n); i++)

int main() {
 multiset<UL> S;
 UL Q; scanf("%u",&Q);
 rep(q,Q){
  UL c; scanf("%u",&c);
  if(c==0){
   UL x; scanf("%u",&x);
   S.insert(x);
   printf("%u\n",S.size());
  }
  if(c==1){
   UL x; scanf("%u",&x);
   printf("%u\n",S.count(x));
  }
  if(c==2){
   UL x; scanf("%u",&x);
   S.erase(x);
  }
  if(c==3){
   UL L,R; scanf("%u%u",&L,&R);
   for(auto itr=S.lower_bound(L); itr!=S.end(); itr++){
    if(*itr>R) break;
    printf("%u\n",*itr);
   }
  }fflush(stdout);
 }
 return 0;
}