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 Perl6 (rakudo-star 2016.01) (judge verdict: Wrong Answer)

my $people_num = prompt('');
my $data = {};
# for 1..$people_num -> $witness_plus_one {
#     my $testimony_num = prompt('');
#     for 1..$testimony_num {
#         my ($person_plus_one, $is_honest) = prompt('').split(/\s/);
#         $data.{$witness_plus_one-1}.{$person_plus_one-1} = $is_honest;
#     }
# }

# e.g. $people_num == 3:
#      => [(0 0 0) (0 0 1) (0 1 0) (0 1 1) (1 0 0) (1 0 1) (1 1 0) (1 1 1)]
my @assumption = [X] (1, 0) xx $people_num;

my $answer = 0;
for @assumption -> $assum {
    my $has_contradict_testimony = False;
    for $assum.kv -> $witness, $assum_is_honest {
        next unless $assum_is_honest;
        next unless $data.{$witness}:exists;
        $has_contradict_testimony = first {
            my ($person, $is_honest) = $_.kv;
            $assum.[$person] != $is_honest;
        }, $data.{$witness}.list;
        last if ?$has_contradict_testimony;
    }
    unless (?$has_contradict_testimony) {
        $answer = [+] $assum;
        last;
    }
}
say $answer;