'Jeopardy!' is an intellectual game where players answer questions and earn points. Company Q conducts a simplified 'Jeopardy!' tournament among the best IT companies. By a lucky coincidence, the old rivals made it to the finals: company R1 and company R2. 

The finals will have n questions, m of them are auction questions and n - m of them are regular questions. Each question has a price. The price of the i-th question is ai points. During the game the players chose the questions. At that, if the question is an auction, then the player who chose it can change the price if the number of his current points is strictly larger than the price of the question. The new price of the question cannot be less than the original price and cannot be greater than the current number of points of the player who chose the question. The correct answer brings the player the points equal to the price of the question. The wrong answer to the question reduces the number of the player's points by the value of the question price.

The game will go as follows. First, the R2 company selects a question, then the questions are chosen by the one who answered the previous question correctly. If no one answered the question, then the person who chose last chooses again.

All R2 employees support their team. They want to calculate what maximum possible number of points the R2 team can get if luck is on their side during the whole game (they will always be the first to correctly answer questions). Perhaps you are not going to be surprised, but this problem was again entrusted for you to solve.

Input

The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100; m ≤ min(n, 30)) — the total number of questions and the number of auction questions, correspondingly. The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 107) — the prices of the questions. The third line contains m distinct integers bi (1 ≤ bi ≤ n) — the numbers of auction questions. Assume that the questions are numbered from 1 to n.

Output

In the single line, print the answer to the problem — the maximum points the R2 company can get if it plays optimally well. It is guaranteed that the answer fits into the integer 64-bit signed type.

Examples

Input

4 1
1 3 7 5
3


Output

18


Input

3 2
10 3 8
2 3


Output

40


Input

2 2
100 200
1 2


Output

400
#Mamma don't raises quitter.................................................
from collections import deque as de
import math
from math import sqrt as sq
from math import floor as fl
from math import ceil as ce
from sys import stdin, stdout
import re
from collections import Counter as cnt
from functools import  reduce

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 find LCM of two numbers
def lcm(x,y):
   lcm = (x*y)//math.gcd(x,y)
   return lcm

def isPowerOfTwo(n): 
    return (math.ceil(Log2(n)) == math.floor(Log2(n))); 
#to check whether the given sorted sequnce is forming an AP or not....
def checkisap(list):
    d=list[1]-list[0]
    for i in range(2,len(list)):
        temp=list[i]-list[i-1]
        if temp !=d:
            return False
    return True
        
#seive of erathanos
def primes_method5(n):
    out ={}
    sieve = [True] * (n+1)
    for p in range(2, n+1):
        if (sieve[p]):
            out[p]=1
            for i in range(p, n+1, p):
                sieve[i] = False
    return out
#function to get the sum of digits
def getSum(n): 
      
    strr = str(n)
    list_of_number = list(map(int, strr.strip()))
    return sum(list_of_number)


#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()))
def indict():
    dic = {}
    for index, value in enumerate(input().split()):
        dic[int(value)] = int(index)+1
    return dic
def frqdict(): 
    # by default it is for integer input. :)
    dic={}
    for index, value in enumerate(input()):
        if value not in dic:
            dic[value] =1
        else:
            dic[value] +=1
    return dic

#inp = open("input.txt","r")
#out = open("output.txt","w")
#Here we go......................
#practice like your never won
#perform like you never lost
n,m=di()
a=li()
b=indict()
if n-m==0:
    mx=max(a)
    ind=a.index(mx)
    for i in range(n):
        if i!=ind:
            mx+=max(a[i],mx )
    print(mx)
else:
    ans=0
    temp=[]
    for i in range(n):
        if i+1 not in b:
            ans+=a[i]
        else:
            temp.append(a[i])
    temp.sort(reverse=True)
    for ele in temp:
        ans+=max(ele, ans)
    print(ans)
            

    
    
    



            

            


    
        


        
    

            
        
    

                
                
                
        
        
        

            

        





                    
                
            
        

            
    
        
    
    

    
        
    
 

    
        


    
        

