Problem:
Score : 300 points

Problem 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.

Constraints

1≤N≤100

0≤F_{i,j,k}≤1

For every integer i such that 1≤i≤N, there exists at least one pair (j,k) such that F_{i,j,k}=1.

-10^7≤P_{i,j}≤10^7

All input values are integers.

Input

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}

Output

Print the maximum possible profit of Joisino's shop.

Sample Input 1

1
1 1 0 1 0 0 0 1 0 1
3 4 5 6 7 8 9 -2 -3 4 -2

Sample Output 1

8

If her shop is open only during the periods when Shop 1 is opened, the profit will be 8, which is the maximum possible profit.

Sample Input 2

2
1 1 1 1 1 0 0 0 0 0
0 0 0 0 0 1 1 1 1 1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1

Sample Output 2

-2

Note that a shop must be open during at least one period, and the profit may be negative.

Sample Input 3

3
1 1 1 1 1 1 0 0 1 1
0 1 0 1 1 1 1 0 1 0
1 0 1 1 0 1 0 1 0 1
-8 6 -2 -8 -8 4 8 7 -6 2 2
-9 2 0 1 7 -5 0 -2 -6 5 5
6 -6 7 -9 6 -5 8 0 -9 -7 -7

Sample Output 3

23

Solution in Ruby (2.3.3) (judge verdict: Accepted)

@n = gets.to_i
@shop_opens = []
@n.times do 
  @shop_opens << gets.chomp.split.map(&:to_i)
end

@shop_points = []
@n.times do
  @shop_points << gets.chomp.split.map(&:to_i)
end

# 各店の開店してる同じ時間帯に開店した回数ぶんだけポイントになる、これを全店でやったポイントの総数の最大値を答える
# -> 開店する時間帯の決め方は月~金それぞれ午前午後で、とりうるすべての組み合わせは2^10程度なのでこれを全探索する
# 開店/閉店は1と0で表されるのを利用してビット演算してANDとる方法もあるがビット演算を調べるのがめんどくさいので愚直に配列でやろうと思う

@ans = -10000000000

time_period = Array.new(10, 0)

def rec(arr, idx)
  if idx>=10
    # 各店で同じ時間帯に開いてる数から利益を算出して比較アンド更新
    if arr.uniq != [0]
      # 「最低一回は開店する」ので「1度も開店してない」状態以外についてのみ考える
      temp_point = 0
      @n.times do |i|
        match_count = 0
        (0...10).each do |j|
          match_count += arr[j] == @shop_opens[i][j] && arr[j] == 1 ? 1 : 0
        end
        temp_point += @shop_points[i][match_count]
      end
      @ans = [@ans, temp_point].max
    end
  else
    # 探索
    nex_arr = arr.dup
    nex_arr[idx] = 1
    rec(nex_arr, idx+1)
    nex_arr[idx] = 0
    rec(nex_arr, idx+1)
  end
end

rec(time_period, 0)

puts @ans