You have a card deck of n cards, numbered from top to bottom, i. e. the top card has index 1 and bottom card — index n. Each card has its color: the i-th card has color a_i.

You should process q queries. The j-th query is described by integer t_j. For each query you should: 

  * find the highest card in the deck with color t_j, i. e. the card with minimum index; 
  * print the position of the card you found; 
  * take the card and place it on top of the deck. 

Input

The first line contains two integers n and q (2 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ q ≤ 3 ⋅ 10^5) — the number of cards in the deck and the number of queries.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 50) — the colors of cards.

The third line contains q integers t_1, t_2, ..., t_q (1 ≤ t_j ≤ 50) — the query colors. It's guaranteed that queries ask only colors that are present in the deck.

Output

Print q integers — the answers for each query.

Example

Input


7 5
2 1 1 4 3 3 1
3 2 1 1 4


Output


5 2 3 1 5 

Note

Description of the sample: 

  1. the deck is [2, 1, 1, 4, \underline{3}, 3, 1] and the first card with color t_1 = 3 has position 5; 
  2. the deck is [3, \underline{2}, 1, 1, 4, 3, 1] and the first card with color t_2 = 2 has position 2; 
  3. the deck is [2, 3, \underline{1}, 1, 4, 3, 1] and the first card with color t_3 = 1 has position 3; 
  4. the deck is [\underline{1}, 2, 3, 1, 4, 3, 1] and the first card with color t_4 = 1 has position 1; 
  5. the deck is [1, 2, 3, 1, \underline{4}, 3, 1] and the first card with color t_5 = 4 has position 5. 
#####################################   
import atexit, io, sys, collections, math, heapq, fractions, bisect,copy
buffer = io.BytesIO() 
sys.stdout = buffer
@atexit.register 
def write(): sys.__stdout__.write(buffer.getvalue())   
####################################



def build(a):
	t = [0 for _ in range(len(a)<< 1)]
	for i in range(len(a)): t[i+len(a)] = a[i]
	for i in range(len(a)-1, 0, -1):t[i] += t[2*i] + t[2*i +1]
	return t
def update(a,i,v,t):
	delta = v - a[i]
	a[i] += delta 
	i += len(a)
	while(i > 0):
		t[i] += delta
		i >>= 1
def query(l,r,t):
	l += len(t) // 2
	r += len(t) // 2
	ans = 0
	while(l <= r):
		if l % 2: 
			ans += t[l]
			l +=1
		if r % 2 == 0: 
			ans += t[r]
			r -=1
		l >>= 1
		r >>= 1
	return ans

n,q= map(int, raw_input().split())
ais = map(int, raw_input().split())

a = [0 for i in range(2 * n)] + [1 for i in range(n)]
t = build(a)

d = {}
for i,e in enumerate(ais):
	if e not in d: d[e] = 2*n+i
head = 2*n - 1
qis = map(int, raw_input().split())
ans = []
for qi in qis:
	idx = d[qi]
	howmanypeople = query(0, idx , t)
	ans.append(howmanypeople if howmanypeople > 0 else 1)
	if howmanypeople > 0:
		update(a,idx, 0,t)
		update(a, head, 1, t)
		d[qi] = head
		head -= 1
	
print ' '.join(map(str, ans))
