Problem: B: Pivots

problem

Given a permutation of length N, a_1, a_2, ..., a_N, which is a permutation of integers from 1 to N. Also, Q queries are given in order for this permutation. In the i-th query, you have to do the following:

* The value q_i (1 \ leq q_i \ leq N) is given. In the permutation \\ {a_1, a_2, ..., a_N \\}, where L is the permutation on the left side of q_i and R is the permutation on the right side of q_i, the original permutation L \\ q_i \\ R is R \\ Change to q_i \\ L. That is, when q_ {i} = a_j, the permutations \\ {a_1, ..., a_ {j-1}, a_j, a_ {j + 1}, ..., a_N \\} are \\ { Change to a_ {j + 1}, ..., a_N, a_j, a_1, ..., a_ {j-1} \\}.



The permutations L and R may be empty. For example, if L is empty, change q_i \\ R to R \\ q_i. The same is true when R is empty.

Output the permutation after processing these Q queries in order for the given permutation on one line.

Input format


N Q
a_1 a_2 ... a_N
q_1 q_2 ... q_Q


All inputs are integers.

The first row gives the number of elements in the permutation N and the number of queries Q, separated by blanks. In the second line, permutations a_1, a_2, ..., a_N, in which integers from 1 to N are rearranged, are given separated by blanks. The third line gives Q queries, separated by blanks. q_i represents the i-th query.

Constraint

* 1 \ leq N \ leq 10 ^ 5
* 1 \ leq Q \ leq 10 ^ 5
* 1 \ leq a_i \ leq N
* a_i is different
* 1 \ leq q_i \ leq N



Output format

Output the permutation after processing all the queries in order on one line.

Input example 1


5 2
1 5 3 2 4
5 2


Output example 1


4 5 1 2 3

* The first query changes the permutation to \\ {3, 2, 4, 5, 1 \\}.
* The second query changes the permutation to \\ {4, 5, 1, 2, 3 \\}.



Input example 2


5 1
1 2 3 4 5
Five


Output example 2


5 1 2 3 4





Example

Input

5 2
1 5 3 2 4
5 2


Output

4 5 1 2 3

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define GET_MACRO(_1,_2,_3,_4,_5,_6,_7,_8,NAME,...) NAME
#define pr(...) cerr<< GET_MACRO(__VA_ARGS__,pr8,pr7,pr6,pr5,pr4,pr3,pr2,pr1)(__VA_ARGS__) <<endl
#define pr1(a) (#a)<<"="<<(a)<<" "
#define pr2(a,b) pr1(a)<<pr1(b)
#define pr3(a,b,c) pr1(a)<<pr2(b,c)
#define pr4(a,b,c,d) pr1(a)<<pr3(b,c,d)
#define pr5(a,b,c,d,e) pr1(a)<<pr4(b,c,d,e)
#define pr6(a,b,c,d,e,f) pr1(a)<<pr5(b,c,d,e,f)
#define pr7(a,b,c,d,e,f,g) pr1(a)<<pr6(b,c,d,e,f,g)
#define pr8(a,b,c,d,e,f,g,h) pr1(a)<<pr7(b,c,d,e,f,g,h)
using namespace std;
using Int = long long;
using _int = int;
using ll = long long;
using Double = long double;
const Int INF = (1LL<<55)+1e9; // ~ 3.6 * 1e16
const Int mod = (1e9)+7;
const Double EPS = 1e-8;
const Double PI = 6.0 * asin((Double)0.5);
using P = pair<Int,Int>;
using T = tuple<Int,Int,Int>;
template<class T> T Max(T &a,T b){return a=max(a,b);}
template<class T> T Min(T &a,T b){return a=min(a,b);}
template<class T1, class T2> ostream& operator<<(ostream& o,pair<T1,T2> p){return o<<"("<<p.first<<","<<p.second<<")";}
template<class T1, class T2, class T3> ostream& operator<<(ostream& o,tuple<T1,T2,T3> t){
  return o<<"("<<get<0>(t)<<","<<get<1>(t)<<","<<get<2>(t)<<")";}
template<class T1, class T2> istream& operator>>(istream& i,pair<T1,T2> &p){return i>>p.first>>p.second;}
template<class T> ostream& operator<<(ostream& o,vector<T> &a){Int i=0;for(T t:a)o<<(i++?" ":"")<<t;return o;}
template<class T> istream& operator>>(istream& i,vector<T> &a){for(T &t:a)i>>t;return i;}
template<class T>void prArr(T a,string s=" "){Int i=0;for(auto t:a)cout<<(i++?s:"")<<t;cout<<endl;}

struct dat{
  int idx,to, pre;
  
  dat(int idx=-1, int to=-1,int pre = -1):idx(idx),to(to), pre(pre){}
};

void print(int head, int tail, vector<dat> A,vector<int> num){
  vector<int>ans;
  for(int i = head; ; i = A[i].to){
    //pr(i, A[i].pre, A[i].to, A[i].idx, head,tail);
    ans.push_back(num[A[i].idx]);
    if(i == tail) break;;
  }
  cout<<ans<<endl;
  
}

signed main(){
  srand((unsigned)time(NULL));
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(12);

  int n, Q;
  cin >>n>> Q;
  vector<dat> A(n);
  for(int i=0;i<n;i++){
    A[i].idx = i;
    A[i].to = (i+1)%n;
    A[i].pre = (i-1 + n) % n;
  }
  int head = 0, tail = n-1;
  
  vector<int> num(n);
  vector<int> Idx(n);
  cin>>num;
  for(int i=0;i<n;i++) Idx[num[i]-1] = i;

  auto query=[&](int x){

    if(x == tail){
      int nhead = x;
      int ntail = A[x].pre;
      A[x].to = head;
      A[head].pre = x;
      head = nhead;
      tail = ntail;
      return;
    };

    if(x == head){
      int nhead = A[x].to;
      int ntail = x;
      A[tail].to = x;
      A[x].pre = tail;
      head = nhead;
      tail = ntail;
      return;
    }

    int nhead = A[x].to;
    int ntail = A[x].pre;
    A[x].to = head;
    A[head].pre = x;

    A[tail].to = x;
    A[x].pre = tail;
    head = nhead;
    tail = ntail;
  };

  
  while(Q--){
    int q;
    cin>>q;
    query(Idx[q-1]);
    // print(head, tail, A,num);
  }

  print(head, tail, A, num);
  
  
  return 0;
}