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 Pascal (FPC 2.6.2) (judge verdict: Time Limit Exceeded)

var n,x,ans,ma,m2,d,m3,da,nn:int64;
var i,j:longint;
var a,b:array[0..1000] of int64;
var c:array[0..102,0..102] of longint;
function max(a,b:int64):int64;
begin
  if a>b then exit(a);exit(b);
end;
procedure dfs(d:int64);
var i:longint;
var num,d1:int64;
begin
  if d>n then
  begin
    num:=0;d1:=-1;
    for i:=1 to n do begin d1:=max(d1,b[i]);inc(num,c[i,b[i]]);end;
    if d1>0 then ans:=max(ans,num);
    exit;
  end;
  for i:=0 to a[d] do begin b[d]:=i;dfs(d+1);end;
end;
begin
  read(n);
  for i:=1 to n do
    for j:=1 to 10 do begin read(x);if x=1 then inc(a[i]);end;
  {ma:=-2147483647;d:=-1;
  for i:=1 to n do
  begin
    m2:=-2147483647;
    for j:=0 to 10 do
    begin
      read(x);if j=0 then b[i]:=x;
      if j<=a[i] then
      begin
        if x>m2 then begin m2:=x;d:=max(d,j);end;
        if j>0 then
        //if (x>ma) or ((x=ma) and (ma-b[da]<x-b[i])) then
        if (ma-b[da]<=x-b[i]) then begin ma:=x;da:=i;end;
      end;
    end;
    if (da=i) and (m2<ma) then begin inc(ans,ma);d:=233;end
    else inc(ans,m2);
  end;  nn:=-2147483647;
  for i:=1 to n do if b[i]<nn then begin nn:=b[i];da:=i;end;
  if d=0 then begin dec(ans,b[da]);inc(ans,ma);end;}
  ans:=-2147483647;
  for i:=1 to n do for j:=0 to 10 do read(c[i,j]);
  dfs(1);
  writeln(ans);
end.