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

Source: aizu | Difficulty: UNKNOWN_DIFFICULTY


Solutions:
def inpl():
	return list(map(int, input().split()))
(N, _) = inpl()
A = inpl()
if N == 1:
	print(*A)
	exit()
L = [0] * (N + 1)
R = [0] * (N + 1)
for i in range(N - 1):
	R[A[i]] = A[i + 1]
	L[A[i + 1]] = A[i]
lm = A[0]
rm = A[-1]
for q in inpl():
	if q == rm:
		l = L[q]
		R[l] = 0
		L[q] = 0
		R[q] = lm
		L[lm] = q
		lm = q
		rm = l
	elif q == lm:
		r = R[q]
		L[r] = 0
		R[q] = 0
		L[q] = rm
		R[rm] = q
		lm = r
		rm = q
	else:
		(l, r) = (L[q], R[q])
		L[q] = rm
		R[q] = lm
		R[l] = 0
		L[r] = 0
		L[lm] = q
		R[rm] = q
		rm = l
		lm = r
ans = []
while lm != 0:
	ans.append(lm)
	lm = R[lm]
print(*ans)