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;
using ll = long long;
using P = pair<int, int>;
#define PI acos(-1)
#define Mod (int)1000000007
#define INFTY (int)INT_MAX
#define Rep(i, n) for (int i = 0; i < (int)(n); i++)
#define BitRep(i, n) for (int i = 0; i < (int)(1 << n); i++)
#define All(vec) vec.begin(), vec.end()
#define Part(vec, b, e) vec.begin() + b, vec.begin() + e
#define Sum(vec) accumulate(vec.begin(), vec.end(), 0)
#define Sort(vec) sort(vec.begin(), vec.end())
#define Reverse(vec) reverse(vec.begin(), vec.end())
#define Count(vec, x) count(vec.begin(), vec.end(), x)
#define Next_permutation(vec) next_permutation(vec.begin(), vec.end())
#define Find(vec, s) vec.find(s) != vec.end()

int main() {
	multiset<int> MS;
	int q;
	cin >> q;
	Rep(i, q) {
		int n;
		cin >> n;
		if (n == 0) {
			int x;
			cin >> x;
			MS.insert(x);
			cout << MS.size() << endl;
		}
		else if (n == 1) {
			int x;
			cin >> x;
			cout << distance(MS.lower_bound(x), MS.upper_bound(x)) << endl;
		}
		else if (n == 2) {
			int x;
			cin >> x;
			MS.erase(x);
		}
		else if (n == 3) {
			int L, R;
			cin >> L >> R;
			for (auto itr = MS.lower_bound(L); itr != MS.end() && *itr <= R; itr++) {
				cout << *itr << endl;
			}
		}
	}
	return 0;
}
