#include <iostream>
#include <set>

using namespace std;
int main() {
  multiset<int> s;
  int q, x;
  cin >> q; // skip n
  while (cin >> q >> x) {
    if (q == 0) {
      s.insert(x);
      cout << s.size() << endl;
    }
    if (q == 1)
      cout << s.count(x) << endl;
    if (q == 2)
      s.erase(x);
    if (q == 3) {
      int r;
      cin >> r;
      auto li = s.lower_bound(x);
      auto ri = s.lower_bound(r + 1);
      while (li != ri)
        cout << *(li++) << endl;
    }
  }
}

### 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 <iostream>
#include <set>

using namespace std;
int main() {
  multiset<int> s;
  int q, x;
  cin >> q; // skip n
  while (cin >> q >> x) {
    if (q == 0) {
      s.insert(x);
      cout << s.size() << endl;
    }
    if (q == 1)
      cout << s.count(x) << endl;
    if (q == 2)
      s.erase(x);
    if (q == 3) {
      int r;
      cin >> r;
      auto li = s.lower_bound(x);
      auto ri = s.lower_bound(r + 1);
      while (li != ri)
        cout << *(li++) << endl;
    }
  }
}

```