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>
using namespace std;
using Int = long long;
//INSERT ABOVE HERE
signed main(){
  Int n,q;
  cin>>n>>q;
  vector<Int> A(n),B(q);
  for(Int i=0;i<n;i++) cin>>A[i];
  for(Int i=0;i<q;i++) cin>>B[i];
  
  if(n==1){
    cout<<1<<endl;
    return 0;
  }

  vector<Int> L(n+1,0),R(n+1,0);
  for(Int i=0;i+1<n;i++){
    R[A[i]]=A[i+1];
    L[A[i+1]]=A[i];
  }
  Int lm=A[0],rm=A[n-1];

  for(Int i=0;i<q;i++){
    Int x=B[i];
    if(x==rm){
      Int l=L[x];

      R[l]=0;
      L[x]=0;
      R[x]=lm;
      L[lm]=x;

      lm = x;
      rm = l;
    }else if(x==lm){
      Int r=R[x]; 

      L[r]=0;
      R[x]=0;
      L[x]=rm;
      R[rm]=x;
      
      lm = r;
      rm = x;
    }else{
      Int l=L[x],r=R[x];
      L[x]=rm;
      R[x]=lm;

      R[l]=0;
      L[r]=0;

      L[lm]=x;
      R[rm]=x;

      rm=l;
      lm=r;
    }    
  }
  
  vector<Int> ans;
  while(lm!=0){
    ans.emplace_back(lm);
    lm=R[lm];
  }
  for(Int i=0;i<n;i++){
    if(i) cout<<" ";
    cout<<ans[i];
  }
  cout<<endl;
  return 0;
}