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 PYTHON3):

Solution:
from bisect import bisect_left, bisect_right, insort_left
dict = {}
li = []
count = 0
q = int(input())
for i in range(q):
    a = input().split()
    t = int(a[1])
    if a[0] == "0":
        if t not in dict:
            dict[t] = 1
            insort_left(li, t)
        else:
            dict[t] += 1
        count += 1
        print(count)
    elif a[0] == "1":
        print(dict[t] if t in dict else 0)
    elif a[0] == "2":
        if t in dict:
            count -= dict[t]
            dict[t] = 0
    else:
        l = bisect_left(li, int(a[1]))
        r = bisect_right(li, int(a[2]))
        lis = []
        for i in range(l, r):
            lis += [li[i]] * dict[li[i]]
        if lis != []:
            print(*lis, sep="\n")