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
# Solved by QBnewb
# Discretization
# off-line

q = int(input())
s = []
rs = [] # for discretization

# download the input
for i in range(q):
    s.append(list(map(int, input().split())))
    if s[i][0] == 3:
        rs.append(s[i][1])
        rs.append(s[i][2])
    else:
        rs.append(s[i][1])

rs.append(10**9 + 1)
rs = sorted(list(set(rs)))
index = {rs[i]:i for i in range(len(rs))} # discretization
counter = [0 for i in range(len(rs) + 1)] # count the number of every element in set
cnt = 0
for i in range(q):
    op, x = s[i][0], s[i][1]
    idx = index[x]
    if op == 0:
        cnt += 1
        counter[idx] += 1
        print(cnt)
    elif op == 1:
        print(counter[idx])
    elif op == 2:
        cnt -= counter[idx]
        counter[idx] = 0
    else:
        l = idx
        r = index[s[i][2]]
        for j in range(l, r+1):
            for k in range(counter[j]):
                print(rs[j])
