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;

int n,q;
int a[100005];
int f[100005],t[100005];
int s,g;


int main(void){
	scanf("%d%d",&n,&q);
	for(int i=0;i<n;i++){
		scanf("%d",&a[i]);
	}
	s=a[0];
	g=a[n-1];
	for(int i=0;i+1<n;i++){
		t[a[i]]=a[i+1];
	}
	for(int i=0;i+1<n;i++){
		f[a[i+1]]=a[i];
		//printf("%d %d\n",a[i+1],a[i]);
	}
	for(int i=0;i<q;i++){
		int b;
		scanf("%d",&b);
		int ns=t[b];
		int ng=f[b];
		if(ns==0){
			t[b]=s;
			//printf("%d %d\n",f[b],t[b]);
			f[s]=b;
			t[f[b]]=0;
			g=f[b];
			f[b]=0;
			s=b;
			continue;
		}
		if(ng==0){
			f[t[b]]=0;
			t[g]=b;
			f[b]=g;
			s=t[b];
			t[b]=0;
			g=b;
			continue;
		}
		//printf("%d %d\n",ns,ng);
		f[t[b]]=0;
		t[f[b]]=0;
		f[b]=g;
		t[g]=b;
		t[b]=s;
		f[s]=b;
		s=ns;
		g=ng;
	}
	int res=s;
	for(int i=0;i<n;i++){
		printf("%d%c",res,i+1==n?'\n':' ');
		res=t[res];
	}
	return 0;
}