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 = {}
keytbl = []
cnt = 0
q = int(input())
for i in range(q):
    a = list(input().split())
    ki = int(a[1])
    if a[0] == "0":
        if ki not in dict:
            dict[ki] =1
            insort_left(keytbl,ki)
        else:dict[ki] += 1
        cnt += 1
        print(cnt)
    elif a[0] == "1":print(dict[ki] if ki in dict else 0)
    elif a[0] == "2":
        if ki in dict:
            cnt -= dict[ki]
            dict[ki] =0
    else:
        L = bisect_left(keytbl,int(a[1]))
        R = bisect_right(keytbl,int(a[2]),L)
        r = []
        for j in range(L,R):r += [keytbl[j]]*dict[keytbl[j]]
        if r != []:print(*r,sep="\n")