Palo Alto is an unusual city because it is an endless coordinate line. It is also known for the office of Lyft Level 5.

Lyft has become so popular so that it is now used by all m taxi drivers in the city, who every day transport the rest of the city residents — n riders.

Each resident (including taxi drivers) of Palo-Alto lives in its unique location (there is no such pair of residents that their coordinates are the same).

The Lyft system is very clever: when a rider calls a taxi, his call does not go to all taxi drivers, but only to the one that is the closest to that person. If there are multiple ones with the same distance, then to taxi driver with a smaller coordinate is selected.

But one morning the taxi drivers wondered: how many riders are there that would call the given taxi driver if they were the first to order a taxi on that day? In other words, you need to find for each taxi driver i the number a_{i} — the number of riders that would call the i-th taxi driver when all drivers and riders are at their home?

The taxi driver can neither transport himself nor other taxi drivers.

Input

The first line contains two integers n and m (1 ≤ n,m ≤ 10^5) — number of riders and taxi drivers.

The second line contains n + m integers x_1, x_2, …, x_{n+m} (1 ≤ x_1 < x_2 < … < x_{n+m} ≤ 10^9), where x_i is the coordinate where the i-th resident lives. 

The third line contains n + m integers t_1, t_2, …, t_{n+m} (0 ≤ t_i ≤ 1). If t_i = 1, then the i-th resident is a taxi driver, otherwise t_i = 0.

It is guaranteed that the number of i such that t_i = 1 is equal to m.

Output

Print m integers a_1, a_2, …, a_{m}, where a_i is the answer for the i-th taxi driver. The taxi driver has the number i if among all the taxi drivers he lives in the i-th smallest coordinate (see examples for better understanding).

Examples

Input

3 1
1 2 3 10
0 0 1 0


Output

3 

Input

3 2
2 3 4 5 6
1 0 0 0 1


Output

2 1 

Input

1 4
2 4 6 10 15
1 1 1 1 0


Output

0 0 0 1 

Note

In the first example, we have only one taxi driver, which means an order from any of n riders will go to him.

In the second example, the first taxi driver lives at the point with the coordinate 2, and the second one lives at the point with the coordinate 6. Obviously, the nearest taxi driver to the rider who lives on the 3 coordinate is the first one, and to the rider who lives on the coordinate 5 is the second one. The rider who lives on the 4 coordinate has the same distance to the first and the second taxi drivers, but since the first taxi driver has a smaller coordinate, the call from this rider will go to the first taxi driver.

In the third example, we have one rider and the taxi driver nearest to him is the fourth one.
#Concentration and mental toughness are margins of victory.................................................
from collections import deque as de
import math
import re
from collections import Counter as cnt
from functools import reduce
from typing import MutableMapping
from itertools import groupby as gb
from fractions import Fraction as fr
from bisect import bisect_left as bl, bisect_right as br

def factors(n):    
    return set(reduce(list.__add__, 
                ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
    
class My_stack():
    def __init__(self):
        self.data = []
    def my_push(self, x):
        return (self.data.append(x))
    def my_pop(self):
        return (self.data.pop())
    def my_peak(self):
        return (self.data[-1])
    def my_contains(self, x):
        return (self.data.count(x))
    def my_show_all(self):
        return (self.data)
    def isEmpty(self):
      return len(self.data)==0

arrStack = My_stack() 

#decimal to binary   
def decimalToBinary(n): 
    return bin(n).replace("0b", "")
#binary to decimal
def binarytodecimal(n):
    return int(n,2)

def isPrime(n) : 
	if (n <= 1) : 
		return False
	if (n <= 3) : 
		return True

	
	if (n % 2 == 0 or n % 3 == 0) : 
		return False

	i = 5
	while(i * i <= n) : 
		if (n % i == 0 or n % (i + 2) == 0) : 
			return False
		i = i + 6

	return True

def get_prime_factors(number):
    prime_factors = []
    while number % 2 == 0:
        prime_factors.append(2)
        number = number / 2
    for i in range(3, int(math.sqrt(number)) + 1, 2):
        while number % i == 0:
            prime_factors.append(int(i))
            number = number / i

    if number > 2:
        prime_factors.append(int(number))
    return prime_factors

def get_frequency(list):
    dic={}
    for ele in list:
        if ele in dic:
            dic[ele] += 1
        else:
            dic[ele] = 1
    return dic
def Log2(x): 
    return (math.log10(x) / 
            math.log10(2)); 

 
# Function to get product of digits
def getProduct(n):
 
    product = 1
 
    while (n != 0):
        product = product * (n % 10)
        n = n // 10
 
    return product

# function to count consecutive duplicate element in an array
def dupconscount(nums):
    element = []
    freque = []
    if not nums:
        return element
    running_count = 1
    for i in range(len(nums)-1):
        if nums[i] == nums[i+1]:
            running_count += 1
        else:
            freque.append(running_count)
            element.append(nums[i])
            running_count = 1
    freque.append(running_count)
    element.append(nums[i+1])
    return element,freque
        

def isPowerOfTwo(n): 
    return (math.ceil(Log2(n)) == math.floor(Log2(n))); 

#ceil  function gives wrong answer after 10^17 so i have to create my own :)
# because i don't want to doubt on my solution of 900-1000 problem set.
def ceildiv(x,y): 
    return (x+y-1)//y 
  
def di():return map(int, input().split())
def ii():return int(input())
def li():return list(map(int, input().split()))
def si():return list(map(str, input()))

#Here we go......................
#concentration and mental toughness are margins of victory
n,m=di()
resi=li()
conf=li()
if m==1:
    print(n)
else:
    driind=[]
    resiind=[]
    for i in range(n+m):
        if conf[i]==1:
            driind.append(resi[i])
        else:
            resiind.append(resi[i])
    ans ={}
    dic={}
    for i in range(m):
        dic[i]=driind[i]
        ans[driind[i]]=0
        
    for i in range(n):
        ii=bl(driind,resiind[i])
        if ii==0:
            ans[dic[0]]+=1
        elif ii==m:

            ans[dic[m-1]]+=1
          
               
        else:
            if abs(dic[ii]-resiind[i]) > abs (dic[ii-1]-resiind[i]):
                mn=dic[ii-1]
            elif abs(dic[ii]-resiind[i]) < abs (dic[ii-1]-resiind[i]):
                mn=dic[ii] 
            else:
                mn= min (dic[ii], dic[ii-1])    
            ans[mn]+=1
    anslist=[]
    for key, val in ans.items():
        anslist.append([key,val])
    anslist.sort()
    for row in anslist:
        print(row[1],end=" ")

        
    




    





    





    
    
    
    
    

          
            
    

    
    
    
    
    
    
        
        
            
            
            
    
        
        
        

                

        
    



        
            

        

        
    
        
         





                
        
            
        
            

        
        

    

    
    
            
        
    

             
        


    
        

            

        
    
    

    
        
    
    

            
            
            

        
    
        
    
    
    
    

    
            
        
        
    
        
        
        


                
                

    


        
        
    
    
    
        
    
        
    




    

                
    
        


        
    

            
        
    

                
                
                
        
        
        

            

        





                    
                
            
        

            
    
        
    
    

    
        
    
 

    
        


    
        

