#include<bits/stdc++.h>
using namespace std;
using lint = long long;
template<typename T = lint>inline T in() { T x; cin >> x; return x; }
#define INF           1e9
#define INFL          static_cast<lint>(INF)
#define REP(i,n)      for(lint i=0,i##_len=(n);i<i##_len;++i)
#define REP1(i,n)     for(lint i=1,i##_len=(n);i<=i##_len;++i)
#define REPR(i,n)     for(lint i=n-1;i>=0;--i)
#define REPR1(i,n)    for(lint i=n;i>0;--i)
#define REPI(i,ini,n) for(lint i=(ini),i##_len=(n);i<i##_len;++i)
#define REPC(i,obj)   for(auto i:obj)
#define R_UP(a,b)     (((a)+(b)-1)/(b))
#define ALL(obj)      (obj).begin(),(obj).end()
int main()
{
	int q = in();
	multiset<int>s;
	REP(i, q)
	{
		int n = in();
		int x, l, r;
		switch (n)
		{
		case 0:
			x = in();
			s.insert(x);
			cout << s.size() << endl;
			break;
		case 1:
			x = in();
			cout << distance(s.lower_bound(x), s.upper_bound(x)) << endl;
			break;
		case 2:
			x = in();
			s.erase(s.lower_bound(x), s.upper_bound(x));
			break;
		case 3:
			l = in();
			r = in();
			for (auto start = s.lower_bound(l), last = s.upper_bound(r); start != last; start++)cout << *start << endl;
			break;
		}
	}
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
using lint = long long;
template<typename T = lint>inline T in() { T x; cin >> x; return x; }
#define INF           1e9
#define INFL          static_cast<lint>(INF)
#define REP(i,n)      for(lint i=0,i##_len=(n);i<i##_len;++i)
#define REP1(i,n)     for(lint i=1,i##_len=(n);i<=i##_len;++i)
#define REPR(i,n)     for(lint i=n-1;i>=0;--i)
#define REPR1(i,n)    for(lint i=n;i>0;--i)
#define REPI(i,ini,n) for(lint i=(ini),i##_len=(n);i<i##_len;++i)
#define REPC(i,obj)   for(auto i:obj)
#define R_UP(a,b)     (((a)+(b)-1)/(b))
#define ALL(obj)      (obj).begin(),(obj).end()
int main()
{
	int q = in();
	multiset<int>s;
	REP(i, q)
	{
		int n = in();
		int x, l, r;
		switch (n)
		{
		case 0:
			x = in();
			s.insert(x);
			cout << s.size() << endl;
			break;
		case 1:
			x = in();
			cout << distance(s.lower_bound(x), s.upper_bound(x)) << endl;
			break;
		case 2:
			x = in();
			s.erase(s.lower_bound(x), s.upper_bound(x));
			break;
		case 3:
			l = in();
			r = in();
			for (auto start = s.lower_bound(l), last = s.upper_bound(r); start != last; start++)cout << *start << endl;
			break;
		}
	}
}

```