output_description: Print the maximum possible profit of Joisino's shop.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      F_{1,1,1} F_{1,1,2} ... F_{1,5,1} F_{1,5,2}
      :
      F_{N,1,1} F_{N,1,2} ... F_{N,5,1} F_{N,5,2}
      P_{1,0} ... P_{1,10}
      :
      P_{N,0} ... P_{N,10}
attempt: # -*- coding: utf-8 -*-
  
  #############
  # Libraries #
  #############
  
  import sys
  
  input = sys.stdin.readline
  
  import math
  
  # from math import gcd
  import bisect
  from collections import defaultdict
  from collections import deque
  from functools import lru_cache
  
  
  #############
  # Constants #
  #############
  
  MOD = 10**9 + 7
  INF = float("inf")
  
  #############
  # Functions #
  #############
  
  
  ######INPUT######
  def I():
      return int(input().strip())
  
  
  def S():
      return input().strip()
  
  
  def IL():
      return list(map(int, input().split()))
  
  
  def SL():
      return list(map(str, input().split()))
  
  
  def ILs(n):
      return list(int(input()) for _ in range(n))
  
  
  def SLs(n):
      return list(input().strip() for _ in range(n))
  
  
  def ILL(n):
      return [list(map(int, input().split())) for _ in range(n)]
  
  
  def SLL(n):
      return [list(map(str, input().split())) for _ in range(n)]
  
  
  ######OUTPUT######
  def P(arg):
      print(arg)
      return
  
  
  def Y():
      print("Yes")
      return
  
  
  def N():
      print("No")
      return
  
  
  def E():
      exit()
  
  
  def PE(arg):
      print(arg)
      exit()
  
  
  def YE():
      print("Yes")
      exit()
  
  
  def NE():
      print("No")
      exit()
  
  
  #####Shorten#####
  def DD(arg):
      return defaultdict(arg)
  
  
  #####Inverse#####
  def inv(n):
      return pow(n, MOD - 2, MOD)
  
  
  ######Combination######
  kaijo_memo = []
  
  
  def kaijo(n):
      if len(kaijo_memo) > n:
          return kaijo_memo[n]
      if len(kaijo_memo) == 0:
          kaijo_memo.append(1)
      while len(kaijo_memo) <= n:
          kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
      return kaijo_memo[n]
  
  
  gyaku_kaijo_memo = []
  
  
  def gyaku_kaijo(n):
      if len(gyaku_kaijo_memo) > n:
          return gyaku_kaijo_memo[n]
      if len(gyaku_kaijo_memo) == 0:
          gyaku_kaijo_memo.append(1)
      while len(gyaku_kaijo_memo) <= n:
          gyaku_kaijo_memo.append(
              gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo), MOD - 2, MOD) % MOD
          )
      return gyaku_kaijo_memo[n]
  
  
  def nCr(n, r):
      if n == r:
          return 1
      if n < r or r < 0:
          return 0
      ret = 1
      ret = ret * kaijo(n) % MOD
      ret = ret * gyaku_kaijo(r) % MOD
      ret = ret * gyaku_kaijo(n - r) % MOD
      return ret
  
  
  ######Factorization######
  def factorization(n):
      arr = []
      temp = n
      for i in range(2, int(-(-(n**0.5) // 1)) + 1):
          if temp % i == 0:
              cnt = 0
              while temp % i == 0:
                  cnt += 1
                  temp //= i
              arr.append([i, cnt])
      if temp != 1:
          arr.append([temp, 1])
      if arr == []:
          arr.append([n, 1])
      return arr
  
  
  #####MakeDivisors######
  def make_divisors(n):
      divisors = []
      for i in range(1, int(n**0.5) + 1):
          if n % i == 0:
              divisors.append(i)
              if i != n // i:
                  divisors.append(n // i)
      return divisors
  
  
  #####GCD#####
  def gcd(a, b):
      while b:
          a, b = b, a % b
      return a
  
  
  #####LCM#####
  def lcm(a, b):
      return a * b // gcd(a, b)
  
  
  #####BitCount#####
  def count_bit(n):
      count = 0
      while n:
          n &= n - 1
          count += 1
      return count
  
  
  #####ChangeBase#####
  def base_10_to_n(X, n):
      if X // n:
          return base_10_to_n(X // n, n) + [X % n]
      return [X % n]
  
  
  def base_n_to_10(X, n):
      return sum(int(str(X)[-i - 1]) * n**i for i in range(len(str(X))))
  
  
  #####IntLog#####
  def int_log(n, a):
      count = 0
      while n >= a:
          n //= a
          count += 1
      return count
  
  
  #############
  # Main Code #
  #############
  
  N = I()
  F = ILL(N)
  P = ILL(N)
  
  for i in range(N):
      F[i] = base_n_to_10(sum(F[i][j] * 10**j for j in range(10)), 2)
  
  ans = 0
  for b in range(1, 1 << 10):
      rieki = 0
      for i in range(N):
          match = count_bit(F[i] & b)
          rieki += P[i][match]
      ans = max(ans, rieki)
  
  print(ans)
  
problem_description: Statement
  
  Joisino is planning to open a shop in a shopping street.
  
  Each of the five weekdays is divided into two periods, the morning and the
  evening. For each of those ten periods, a shop must be either open during the
  whole period, or closed during the whole period. Naturally, a shop must be
  open during at least one of those periods.
  
  There are already N stores in the street, numbered 1 through N.
  
  You are given information of the business hours of those shops, F_{i,j,k}. If
  F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is
  explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here,
  the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2,
  Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted
  as Period 1, and the afternoon is denoted as Period 2.
  
  Let c_i be the number of periods during which both Shop i and Joisino's shop
  are open. Then, the profit of Joisino's shop will be
  P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}.
  
  Find the maximum possible profit of Joisino's shop when she decides whether
  her shop is open during each period, making sure that it is open during at
  least one period.
samples: [{"input": "1\n    1 1 0 1 0 0 0 1 0 1\n    3 4 5 6 7 8 9 -2 -3 4 -2", "output": "8\n    \n\nIf her shop is open only during the periods when Shop 1 is opened, the profit\nwill be 8, which is the maximum possible profit.\n\n* * *"}, {"input": "2\n    1 1 1 1 1 0 0 0 0 0\n    0 0 0 0 0 1 1 1 1 1\n    0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1\n    0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1", "output": "-2\n    \n\nNote that a shop must be open during at least one period, and the profit may\nbe negative.\n\n* * *"}, {"input": "3\n    1 1 1 1 1 1 0 0 1 1\n    0 1 0 1 1 1 1 0 1 0\n    1 0 1 1 0 1 0 1 0 1\n    -8 6 -2 -8 -8 4 8 7 -6 2 2\n    -9 2 0 1 7 -5 0 -2 -6 5 5\n    6 -6 7 -9 6 -5 8 0 -9 -7 -7", "output": "23"}]