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
//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
#define rep(i, n) for(int i=0; i<n; ++i)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
using namespace std;
using ll = int64_t;
using ld = long double;
using P = pair<int, int>;
using vs = vector<string>;
using vi = vector<int>;
using vvi = vector<vi>;
const int INF = 100010001;
const ll LINF = (ll)INF*INF*10;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int q;
    cin >> q;
    set<int> se;
    map<int, int> mp;
    int sum = 0;
    while(q--) {
        int t, x;
        cin >> t >> x;
        if(t == 0) {
            se.insert(x);
            sum++;
            cout << sum << '\n';
            mp[x]++;
        } else if(t == 1) {
            set<int>::iterator itr = se.find(x);
            cout << (itr != se.end()?mp[x]:0) << '\n';
        } else if(t == 2) {
            sum -= mp[x];
            se.erase(x);
            mp.erase(x);
        } else {
            int r;
            cin >> r;
            auto itr = se.lower_bound(x);
            for(;*itr <= r and itr != se.end(); itr = next(itr)) {
                rep(i, mp[*itr]) {
                    cout << *itr << '\n';
                }
            }
        }
    }
}
