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: # https://atcoder.jp/contests/abc080/tasks/abc080_c
  
  # 各店の各曜日の各営業時間で営業するかどうか全探索
  
  
  N = int(input())  # 店の数
  F_list = []
  for _ in range(N):
      f = list(map(int, input().split()))
      F_list.append(f)
  
  P_list = []
  for _ in range(N):
      p = list(map(int, input().split()))
      P_list.append(p)
  
  # # 各曜日時間帯の最低限の店の開店数（おねーちゃんの店を考慮しない）
  # min_open_list = []
  # for i in range(10):
  #     count_open = 0
  #     for j in range(N):
  #         if F_list[j][i] == 1:
  #             count_open += 1
  #     min_open_list.append(count_open)
  
  
  # c個の店が営業中の時の、店iの利益
  def benefit(i, c):
      return P_list[i][c]
  
  
  is_joisino_shop_open_list = [False] * 10
  benefit_joisino_list = []
  
  
  def dfs(j):
      if j >= 10:
  
          # 1つ以上の時間帯でお店を営業していること
          if True in is_joisino_shop_open_list:
              # joisinoおねえちゃんのお店の利益を計算
              benefit_joisino = 0
              for n in range(N):
                  # 各お店ごとのjoisinoお姉ちゃんの利益を計算
                  count_n_open = 0
                  shop_n_list = F_list[n]
  
                  for i in range(10):
                      if shop_n_list[i] == 1 and is_joisino_shop_open_list[i] == True:
                          count_n_open += 1
  
                  benefit_n_joisino = P_list[n][count_n_open]
                  benefit_joisino += benefit_n_joisino
  
              benefit_joisino_list.append(benefit_joisino)
  
          return
  
      # joisinoお姉ちゃんのお店をあけるかあけないか全パターン試す
      # j番目の日時にjoisinoお姉ちゃんのお店をあけるとき
      is_joisino_shop_open_list[j] = True
      dfs(j + 1)
  
      # j番目の日時にjoisinoお姉ちゃんのお店をあけないとき
      is_joisino_shop_open_list[j] = False
      dfs(j + 1)
  
  
  dfs(0)
  max_benefit_joisino = max(benefit_joisino_list)
  print(max_benefit_joisino)
  
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"}]