Problem:
Score : 300 points

Problem Statement

There are N people numbered 1 to N. Each of them is either an honest person whose testimonies are always correct or an unkind person whose testimonies may be correct or not.

Person i gives A_i testimonies. The j-th testimony by Person i is represented by two integers x_{ij} and y_{ij}. If y_{ij} = 1, the testimony says Person x_{ij} is honest; if y_{ij} = 0, it says Person x_{ij} is unkind.

How many honest persons can be among those N people at most?

Constraints

All values in input are integers.

1 \leq N \leq 15

0 \leq A_i \leq N - 1

1 \leq x_{ij} \leq N

x_{ij} \neq i

x_{ij_1} \neq x_{ij_2} (j_1 \neq j_2)

y_{ij} = 0, 1

Input

Input is given from Standard Input in the following format:

N
A_1
x_{11} y_{11}
x_{12} y_{12}
:
x_{1A_1} y_{1A_1}
A_2
x_{21} y_{21}
x_{22} y_{22}
:
x_{2A_2} y_{2A_2}
:
A_N
x_{N1} y_{N1}
x_{N2} y_{N2}
:
x_{NA_N} y_{NA_N}

Output

Print the maximum possible number of honest persons among the N people.

Sample Input 1

3
1
2 1
1
1 1
1
2 0

Sample Output 1

2

If Person 1 and Person 2 are honest and Person 3 is unkind, we have two honest persons without inconsistencies, which is the maximum possible number of honest persons.

Sample Input 2

3
2
2 1
3 0
2
3 1
1 0
2
1 1
2 0

Sample Output 2

0

Assuming that one or more of them are honest immediately leads to a contradiction.

Sample Input 3

2
1
2 0
1
1 0

Sample Output 3

1

Solution in OCaml (4.02.3) (judge verdict: Accepted)

let rec pow x n = if n = 1 then x else x * pow x (n - 1)

let n = Scanf.sscanf (read_line ()) "%d" @@ fun n -> n
let a = Array.init n @@ fun _ ->
  let l = Scanf.sscanf (read_line ()) "%d" @@ fun l -> l in
  Array.init l @@ fun _ -> Scanf.sscanf (read_line ()) "%d %d" @@ fun x y -> (x - 1, y)

let x = pow 2 n

let rec to_list x = function
  | 0 -> []
  | n -> x mod 2 :: to_list (x / 2) (n - 1)

let f arr = 
  let res = Array.mapi(fun i t ->
    Array.fold_left (fun z (x, y) -> 
      z && (t = 0 || arr.(x) = y)
    ) true a.(i)
  ) arr in
  Array.fold_left (&&) true res

let rec loop m i = 
  if i = x then m 
  else begin
    let arr = Array.of_list @@ to_list i n in
    loop (if f arr then max m (Array.fold_left (+) 0 arr) else m) (i + 1)
  end

let () = Printf.printf "%d\n" @@ loop 0 0