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 <iostream>
#include <vector>
using namespace std;

struct node {
    int val;
    node *left, *right;
    node() : val(-1), left(NULL), right(NULL) { }
};

void print(node* cur) {
    while (cur != NULL) {
        cout << cur->val + 1;
        if (cur->right) cout << " ";
        cur = cur->right;
    }
    cout << endl;
}

int main() {
    int N, Q; cin >> N >> Q;
    vector<int> a(N);
    for (int i = 0; i < N; ++i) cin >> a[i], --a[i];

    vector<node*> vec(N);
    node *head = new node();
    node *tail = new node();
    for (int i = 0; i < N; ++i) vec[i] = new node();
    for (int i = 0; i < N; ++i) {
        if (i > 0) vec[a[i]]->left = vec[a[i-1]];
        else head = vec[a[i]];
        if (i+1 < N) vec[a[i]]->right = vec[a[i+1]];
        else tail = vec[a[i]];
        vec[a[i]]->val = a[i];
    }

    //print(head);

    for (int _ = 0; _ < Q; ++_) {
        int q; cin >> q; --q;

        if (vec[q] == head) {
            vec[q]->left = tail;
            tail->right = vec[q];
            vec[q]->right->left = NULL;
            head = vec[q]->right;
            vec[q]->right = NULL;
            tail = vec[q];
        }
        else if (vec[q] == tail) {
            vec[q]->right = head;
            head->left = vec[q];
            vec[q]->left->right = NULL;
            tail = vec[q]->left;
            vec[q]->left = NULL;
            head = vec[q];
        }
        else {
            vec[q]->left->right = NULL;
            swap(vec[q]->left, tail);
            vec[q]->left->right = vec[q];
            vec[q]->right->left = NULL;
            swap(vec[q]->right, head);
            vec[q]->right->left = vec[q];
        }
        
        //print(head);
    }

    print(head);
}