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;
typedef long long LL;

#define FOR(i,bg,ed) for(int i=(bg);i<(ed);i++)
#define REP(i,n) FOR(i,0,n)

#define ALL(v) (v).begin(),(v).end()
#define REC(ret, ...) std::function<ret (__VA_ARGS__)>
template <typename T>inline bool chmin(T &l, T r){bool a = l>r; if (a)l = r; return a;}
template <typename T>inline bool chmax(T &l, T r){bool a = l<r; if (a)l = r; return a;}


int a[1123456];
int pos[1123456];
int l[1123456];
int r[1123456];
int main() {
	int N, Q;
	cin >> N >> Q;
	REP(i, N) {
		cin >> a[i];
		pos[a[i]] = i;
	}
	int lmost = a[0];
	int rmost = a[N - 1];
	l[a[0]] = -1;
	r[a[N - 1]] = -1;
	REP(i, N - 1) {
		r[a[i]] = a[i + 1];
	}
	FOR(i, 1, N) {
		l[a[i]] = a[i - 1];
	}
	while (Q--) {
		int p;
		cin >> p;
		int next_lmost = r[p];
		if (next_lmost == -1)next_lmost = p;
		int next_rmost = l[p];
		if (next_rmost == -1)next_rmost = p;
		if (rmost != p) {
			r[rmost] = p;
			l[r[p]] = -1;
		}
		if (lmost != p) {
			l[lmost] = p;
			r[l[p]] = -1;
		}
		

		if (lmost == p) {
			r[p] = -1;
		}else {
			r[p] = lmost;
		}
		
		if (rmost == p) {
			l[p] = -1;
		}else {
			l[p] = rmost;
		}

		lmost = next_lmost;
		rmost = next_rmost;
	}
	string sp(N, ' ');
	sp[N-1] = '\n';

	
	REP(i, N) {
		cout << lmost << sp[i];
		lmost = r[lmost];
	}
	return 0;
}