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 () =
    let main () =
        let n = int_of_string (read_line ()) in
        let xy = Array.make_matrix n n (-1) in
        for i = 0 to n - 1 do
            let a1 = int_of_string (read_line ()) in
            for j = 0 to a1 - 1 do
                Scanf.sscanf (read_line ()) "%d %d" (fun x y -> xy.(i).(x - 1) <- y)
            done
        done;
        let bits a = 
            let rec loop a acc =
                if a = 0 then acc else loop (a lsr 1) (acc + (a land 1))
            in
            loop a 0
        in
        let judge i =
            let judge2 t =
                if i land (1 lsl t) = 0 then true else (
                    let rec loop2 k =
                        if k = n then true else (
                            if xy.(t).(k) < 0 then loop2 (k + 1) else (
                                if (xy.(t).(k) = 0 && (i land (1 lsl k) = 0)) ||
                                   (xy.(t).(k) = 1 && (i land (1 lsl k) != 0)) then loop2 (k + 1)
                                else false
                            )
                        )
                    in
                    loop2 0
                )
            in
            let rec loop s =
                if s = n then true else (
                    if judge2 s then loop (s + 1) else false
                )
            in
            loop 0
        in
        let maxn = 1 lsl n in
        let rec loop i acc =
            if i = maxn then acc else
                let acc = if judge i then max acc (bits i) else acc in
                loop (i + 1) acc
        in
        let r = loop 0 0 in
        Printf.printf "%d\n" r
    in
    main ()