Problem: For a dynamic list $L$ of integers, perform a sequence of the following operations. $L$ has a special element called END at the end of the list and an element of $L$ is indicated by a cursor.

* insert($x$): Insert $x$ before the element indicated by the cursor. After this operation, the cursor points the inserted element.
* move($d$): Move the cursor to the end by $d$, if $d$ is positive. Move the cursor to the front by $d$, if $d$ is negative.
* erase(): Delete the element indicated by the cursor. After this operation, the cursor points the element next to the deleted element. In case there is no such element, the cursor should point END.



In the initial state, $L$ is empty and the cursor points END.

Constraints

* $1 \leq q \leq 500,000$
* The cursor indicates an element of $L$ or END during the operations
* Erase operation will not given when the cursor points END
* $-1,000,000,000 \leq x \leq 1,000,000,000$
* Moving distance of the cursor ($\sum{|d|}$) does not exceed 1,000,000
* $L$ is not empty after performing all operations

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 $d$


or


2


where the first digits 0, 1 and 2 represent insert, move and erase operations respectively.

Output

Print all elements of the list in order after performing given operations. Print an element in a line.

Example

Input

5
0 1
0 2
0 3
1 1
2


Output

3
1

Solution (in CPP):

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

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef vector<ld> vd;
typedef vector<ll> vll;
typedef vector<string> vstr;
typedef vector<vll> vvll;
typedef vector<vvll> vvvll;
typedef vector<pair<ll, ll>> 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 FOR(i, a, b) for (ll i = (ll)a; i < (ll)b; i++)
#define IREP(i, v) for (auto i = (v).begin(); i != (v).end(); ++i)
#define FI first
#define SE second
#define MP make_pair
#define MT make_tuple
#define PB push_back
#define TS to_string
#define NP next_permutation
#define ALL(v) (v).begin(), (v).end()
#define SZ(x) (ll) x.size()
#define SP(x) setprecision((ll)x)


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);

  list<ll> lis;
  ll q;
  cin >> q;
  auto itr = lis.end();

  REP(i,q){
    ll com;
    cin >> com;
    if(com == 0){
      ll x;
      cin >> x;
      itr = lis.insert(itr,x);
    }
    else if(com == 1){
      ll d;
      cin >> d;
      if(d >= 0) for(ll j=1;j<=d;j++) ++itr;
      else for(ll j=1;j<=abs(d);j++) --itr;
    }
    else{
      itr = lis.erase(itr);
    }
  }

  IREP(i,lis) cout << (*i) <<endl;


  return 0;
}