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;
#define rep(i,n) for(int i=0;i<int(n);i++)
#define rep1(i,n) for(int i=1;i<=int(n);i++)

#define debug(x) cerr<<#x<<": "<<x<<endl

int main(){
    int n,Q;
    cin>>n>>Q;
    vector<int> a(n+2), pre(n+2), nxt(n+2), q(Q);
    rep1(i,n){
        cin>>a[i];
    }
    rep(i,Q) cin>>q[i];
    a[0] = 0;
    a[n+1] = n+1;
    rep1(i,n){
        pre[a[i]] = a[i-1];
        nxt[a[i]] = a[i+1];
    }
    nxt[0] = a[1];
    pre[n+1] = a[n];

    rep(j,Q){
        int p = q[j];
        int v1 = nxt[0];
        int v2 = pre[p];
        int v3 = nxt[p];
        int v4 = pre[n+1];


        if(v3 == n+1){
            nxt[0] = p;
            pre[p] = 0;
        }else{
            nxt[0] = v3;
            pre[v3] = 0;

            nxt[v4] = p;
            pre[p] = v4;
        }

        if(v2 == 0){
            pre[n+1] = p;
            nxt[p] = n+1;
        }else{
            nxt[p] = v1;
            pre[v1] = p;

            nxt[v2] = n+1;
            pre[n+1] = v2;
        }

    }

    int cur = nxt[0];
    int i = 0;
    while(cur != n+1){
        cout<<cur<<(i < n-1 ? " " : "\n");
        cur = nxt[cur];
        i++;
    }
}