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 CPP):

Solution:
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<vvll> vvvll;
typedef vector<ld> vld;
typedef vector<string> vstr;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef priority_queue<ll, vector<ll>, greater<ll>> spqll;  // 小さい順に取り出し
typedef priority_queue<ll, vector<ll>, less<ll>> bpqll;     // 大きい順に取り出し

#define REP(i, n) for (ll i = 0; i < (ll)(n); i++)
#define IREP(i, v) for (auto i = (v).begin(); i != (v).end(); ++i)
#define TS to_string
#define ALL(v) (v).begin(), (v).end()
#define endl "\n"

ll INF = 1e9;
ll MOD = 1000000007;
ll LINF = 1e18;
ld EPS = 1e-9;
ld PI = M_PI;
vll dx = {1, 0, -1, 0, 1, -1, -1, 1};
vll dy = {0, 1, 0, -1, 1, 1, -1, -1};
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a;} //最大公約数
ll lcm(ll a, ll b) { return a / gcd(a, b) * b;}     //最小公倍数

void yes(){ cout << "Yes" << endl;}
void no(){ cout << "No" << endl;}

//-----------------------------------------

//-----------------------------------------
int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);

  ll q;
  cin >> q;
  multiset<ll> st;
  REP(i,q){
    ll com;
    cin >> com;
    if(com == 0){
      ll x; cin >> x;
      st.insert(x);
      cout << st.size() <<endl;
    }
    else if(com == 1){
      ll x; cin >> x;
      auto s = st.lower_bound(x);
      auto t = st.upper_bound(x);
      cout << distance(s,t) << endl;
    }
    else if(com == 2){
      ll x;
      cin >> x;
      st.erase(x);
    }
    else{
      ll l, r;
      cin >> l >> r;
      auto s = st.lower_bound(l);
      auto t = st.upper_bound(r);
      for(auto i = s; i != t; ++i){
        cout << (*i) << endl;
      }
    }
  }
  return 0;
}