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 <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <set>
#define long long long
#define LF '\n'
#define ALL(x) x.begin(), x.end()
using namespace std;
typedef pair<int,int> pii;
template<class A, class B>inline bool chmax(A &a, const B &b){return b>a ? a=b,1 : 0;}
template<class A, class B>inline bool chmin(A &a, const B &b){return b<a ? a=b,1 : 0;}
constexpr int INF = 0x3f3f3f3f;


signed main()
{
  cin.tie(nullptr), ios::sync_with_stdio(false);
  multiset<int> U;
  int Q;

  cin >> Q;
  while(Q--) {
    int com, x, y;
    cin >> com;

    switch(com) {
    case 0:
      cin >> x;
      U.insert(x);
      cout << U.size() << LF;
      break;

    case 1:
      cin >> x;
      cout << U.count(x) << endl;
      break;

    case 2:
      cin >> x;
      U.erase(x);
      break;

    default:
      cin >> x >> y;
      for (auto it = U.lower_bound(x), end=U.upper_bound(y); it != end; ++it) {
        cout << *it << LF;
      }
    }
  }

  return 0;
}