#include <vector>
#include <iostream>
#include <set>

int main() {
	int n;
	std::cin >> n;
	std::multiset<int> set;
	for (auto i = 0; i < n; ++i) {
		int o, x, r;
		std::cin >> o >> x;
		switch (o) {
		case 0: set.insert(x); std::cout << set.size() << std::endl; break;
		case 1: std::cout << set.count(x) << std::endl; break;
		case 2: set.erase(x); break;
		default:
			std::cin >> r;
			auto end = set.upper_bound(r);
			for (auto begin = set.lower_bound(x); begin != end; ++begin) {
				std::cout << *begin << std::endl;
			}
		}
	}
}

### Prompt

Please create a solution in CPP to the following 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

### Response

```cpp
#include <vector>
#include <iostream>
#include <set>

int main() {
	int n;
	std::cin >> n;
	std::multiset<int> set;
	for (auto i = 0; i < n; ++i) {
		int o, x, r;
		std::cin >> o >> x;
		switch (o) {
		case 0: set.insert(x); std::cout << set.size() << std::endl; break;
		case 1: std::cout << set.count(x) << std::endl; break;
		case 2: set.erase(x); break;
		default:
			std::cin >> r;
			auto end = set.upper_bound(r);
			for (auto begin = set.lower_bound(x); begin != end; ++begin) {
				std::cout << *begin << std::endl;
			}
		}
	}
}

```