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;

int main() {
  ios::sync_with_stdio(false);

  int n;
  cin >> n;

  int cmd, val;
  int lhd, rhd;

  multiset<int> s;
  multiset<int>::iterator it;
  for (int i = 0; i < n; i++) {
    cin >> cmd;
    switch (cmd) {
      case 0:
        cin >> val;
        s.insert(val);
        cout << s.size() << endl;
        break;

      case 1:
        cin >> val;
        cout << s.count(val) << endl;
        break;

      case 2:
        cin >> val;
        s.erase(val);
        break;

      case 3:
        cin >> lhd >> rhd;

        for (auto itr = s.lower_bound(lhd); itr != s.end() && *itr <= rhd;
             itr++) {
          cout << *itr << endl;
        }
        break;
    }
  }
}