human
Consider a tunnel on a one-way road. During a particular day, n cars numbered from 1 to n entered and exited the tunnel exactly once. All the cars passed through the tunnel at constant speeds.

A traffic enforcement camera is mounted at the tunnel entrance. Another traffic enforcement camera is mounted at the tunnel exit. Perfectly balanced.

Thanks to the cameras, the order in which the cars entered and exited the tunnel is known. No two cars entered or exited at the same time.

Traffic regulations prohibit overtaking inside the tunnel. If car i overtakes any other car j inside the tunnel, car i must be fined. However, each car can be fined at most once.

Formally, let's say that car i definitely overtook car j if car i entered the tunnel later than car j and exited the tunnel earlier than car j. Then, car i must be fined if and only if it definitely overtook at least one other car.

Find the number of cars that must be fined. 

Input

The first line contains a single integer n (2 ≤ n ≤ 10^5), denoting the number of cars.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), denoting the ids of cars in order of entering the tunnel. All a_i are pairwise distinct.

The third line contains n integers b_1, b_2, …, b_n (1 ≤ b_i ≤ n), denoting the ids of cars in order of exiting the tunnel. All b_i are pairwise distinct.

Output

Output the number of cars to be fined.

Examples

Input

5
3 5 2 1 4
4 3 2 5 1

Output

2

Input

7
5 2 3 6 7 1 4
2 3 6 7 1 4 5

Output

6

Input

2
1 2
1 2

Output

0

Note

The first example is depicted below:


Car 2 definitely overtook car 5, while car 4 definitely overtook cars 1, 2, 3 and 5. Cars 2 and 4 must be fined.

In the second example car 5 was definitely overtaken by all other cars.

In the third example no car must be fined.
Using python can you solve the prior task?
assistant
#Here is Python program to find the number of cars that must be fined.

mod=10**9+7
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
#import threading
#threading.stack_size(2**27)
#import sys
#sys.setrecursionlimit(10**6)
#fact=[1]
#for i in range(1,1000001):
#    fact.append((fact[-1]*i)%mod)
#ifact=[0]*1000001
#ifact[1000000]=pow(fact[1000000],mod-2,mod)
#for i in range(1000000,0,-1):
#    ifact[i-1]=(i*ifact[i])%mod
from sys import stdin, stdout
import bisect
from bisect import bisect_left as bl              #c++ lowerbound bl(array,element)
from bisect import bisect_right as br             #c++ upperbound
import itertools
import collections
import math
import heapq
from random import randint as rn
from Queue import Queue as Q
def modinv(n,p):
    return pow(n,p-2,p)
def ncr(n,r,p):                        #for using this uncomment the lines calculating fact and ifact
    t=((fact[n])*((ifact[r]*ifact[n-r])%p))%p
    return t
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input().strip()
def GCD(x,y):
    while(y):
        x, y = y, x % y
    return x
"""*******************************************************"""
def main():
    n=int(input())
    a=ain()
    b=ain()
    f=[0]*(n+1)
    d={}
    t=1
    for i in range(n):
        d[a[i]]=t
        t+=1
    for i in range(n):
        b[i]=d[b[i]]
    j=0
    s=0
    for i in range(n):
        if((j+1)!=b[i]):
            s+=1
            f[b[i]]=1
        else:
            f[j+1]=1
            while(j<n and f[j+1]==1):
                j+=1
    print s
######## Python 2 and 3 footer by Pajenegod and c1729
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write...