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, head, last;
int pre[100005] = {0};
int suf[100005] = {0};

void solve(int x);

int main() {
  int bf;
  cin >> n >> q;
  if(n == 1) {
    cout << 1 << endl;
    return 0;
  }
  for(int i = 0; i < n; ++i) {
    int x;
    cin >> x;
    if(i == 0) {
      head = x;
      pre[x] = -1;
    }
    else {
      if(i == n - 1) {
        last = x;
        suf[x] = -1;
      }
      pre[x] = bf;
      suf[bf] = x;
    }
    bf = x;
  }
  for(int i = 0; i < q; ++i) {
    int x;
    cin >> x;
    solve(x);
  }
  while(head != -1) {
    cout << head;
    if(suf[head] != -1) cout << " ";
    head = suf[head];
  }
  cout << endl;
  return 0;
}
void solve(int x) {
  if(x == head) {
    head = suf[x];
    suf[last] = x;
    suf[x] = -1;
    pre[head] = -1;
    pre[x] = last;
    last = x;
    return;
  }
  if(x == last) {
    last = pre[x];
    pre[head] = x;
    pre[x] = -1;
    suf[last] = -1;
    suf[x] = head;
    head = x;
    return;
  }
  int nhead = suf[x], nlast = pre[x];
  pre[nhead] = -1;
  suf[nlast] = -1;
  suf[x] = head;
  pre[x] = last;
  suf[last] = x;
  pre[head] = x;
  head = nhead;
  last = nlast;
}