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 "bits/stdc++.h"
using namespace std;
#define long int64_t

struct Solver {
	Solver() {
		ios::sync_with_stdio( false );
		cout << fixed << setprecision( 10 );
	}

	void solve() {
		int q;
		cin >> q;

		multiset<int> s;
		while( q-- ) {
			int cmd, x, y;
			cin >> cmd;

			if( cmd == 0 ) {
				cin >> x;
				s.insert( x );
				cout << s.size() << '\n';
			}
			else if( cmd == 1 ) {
				cin >> x;
				cout << s.count( x ) << '\n';
			}
			else if( cmd == 2 ) {
				cin >> x;
				s.erase( x );
			}
			else {
				cin >> x >> y;
				auto a = s.lower_bound( x );
				auto b = s.upper_bound( y );
				while( a != b ) {
					cout << *a << '\n';
					++a;
				}
			}
		}
	}
};

int main() {
	Solver().solve();
	return 0;
}

