Vasya has got n books, numbered from 1 to n, arranged in a stack. The topmost book has number a_1, the next one — a_2, and so on. The book at the bottom of the stack has number a_n. All numbers are distinct.

Vasya wants to move all the books to his backpack in n steps. During i-th step he wants to move the book number b_i into his backpack. If the book with number b_i is in the stack, he takes this book and all the books above the book b_i, and puts them into the backpack; otherwise he does nothing and begins the next step. For example, if books are arranged in the order [1, 2, 3] (book 1 is the topmost), and Vasya moves the books in the order [2, 1, 3], then during the first step he will move two books (1 and 2), during the second step he will do nothing (since book 1 is already in the backpack), and during the third step — one book (the book number 3). Note that b_1, b_2, ..., b_n are distinct.

Help Vasya! Tell him the number of books he will put into his backpack during each step.

Input

The first line contains one integer n~(1 ≤ n ≤ 2 ⋅ 10^5) — the number of books in the stack.

The second line contains n integers a_1, a_2, ..., a_n~(1 ≤ a_i ≤ n) denoting the stack of books.

The third line contains n integers b_1, b_2, ..., b_n~(1 ≤ b_i ≤ n) denoting the steps Vasya is going to perform.

All numbers a_1 ... a_n are distinct, the same goes for b_1 ... b_n.

Output

Print n integers. The i-th of them should be equal to the number of books Vasya moves to his backpack during the i-th step.

Examples

Input

3
1 2 3
2 1 3


Output

2 0 1 


Input

5
3 1 4 2 5
4 5 1 3 2


Output

3 2 0 0 0 


Input

6
6 5 4 3 2 1
6 5 3 4 2 1


Output

1 1 2 0 1 1 

Note

The first example is described in the statement.

In the second example, during the first step Vasya will move the books [3, 1, 4]. After that only books 2 and 5 remain in the stack (2 is above 5). During the second step Vasya will take the books 2 and 5. After that the stack becomes empty, so during next steps Vasya won't move any books.
import sys
import math
import itertools
import functools
import collections
import operator


def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def revn(n): return str(n)[::-1]
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)
def sieve(n):
    if n < 2: return list()
    prime = [True for _ in range(n + 1)]
    p = 3
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 2
    r = [2]
    for p in range(3, n + 1, 2):
        if prime[p]:
            r.append(p)
    return r
def divs(n, start=2):
    r = []
    for i in range(start, int(math.sqrt(n) + 1)):
        if (n % i == 0):
            if (n / i == i):
                r.append(i)
            else:
                r.extend([i, n // i])
    return r
def divn(n, primes):
    divs_number = 1
    for i in primes:
        if n == 1:
            return divs_number
        t = 1
        while n % i == 0:
            t += 1
            n //= i
        divs_number *= t
def prime(n):
    if n == 2: return True
    if n % 2 == 0 or n <= 1: return False
    sqr = int(math.sqrt(n)) + 1
    for d in range(3, sqr, 2):
        if n % d == 0: return False
    return True
def convn(number, base):
    newnumber = 0
    while number > 0:
        newnumber += number % base
        number //= base
    return newnumber
def cdiv(n, k): return n // k + (n % k != 0)


n = ii()
a = li()
b = li()
ans = []
s = set()
pos = 0
for i in range(n):
    if b[i] in s:
        ans.append(0)
        continue
    else:
        c = 0
        while b[i] != a[pos]:
            s.add(a[pos])
            pos += 1
            c += 1
        else:
            s.add(a[pos])
            pos += 1
            c += 1
        ans.append(c)
print(*ans)
