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 Ruby (2.7.1) (judge verdict: Wrong Answer)

N=gets.to_i
HONESTS={}
UNKINDS={}
(1..N).each do |n|
    honests = [n]
    unkinds = []
    a=gets.to_i
    a.times do
        x,y=gets.chomp.split(" ").map(&:to_i)
        if y == 1
            honests.push(x)
        else
            unkinds.push(x)
        end
    end
    HONESTS[n] = honests
    UNKINDS[n] = unkinds
end
(1..N).each do |n|
    honests = HONESTS[n]
    plus = HONESTS[n]
    while !plus.empty? do
        next_plus = []
        plus.map {|h| HONESTS[h] }.flatten.uniq.each do |h|
            next_plus.push(h) if !honests.include?(h)
        end
        honests.concat(next_plus)
        plus = next_plus
    end
    HONESTS[n] = honests
end

all = (1..N).to_a
unkinds = []
(1..N).each do |n|
    if !(HONESTS[n] & UNKINDS[n]).empty?
        all.delete(n)
        unkinds.push(n)
    end
end

def calc(honests, unkinds, unknowns)
    return honests.length if unknowns.empty?
    ret = honests.length
    unknowns.each do |n|
        honests_by_n = HONESTS[n]
        unkinds_by_n = UNKINDS[n]
        if honests.all? { |h| !unkinds_by_n.include?(h) } && unkinds.all? { |u| !honests_by_n.include?(u) }
            next_unkown = unknowns.dup
            next_unkown.delete(n)
            r = calc(
                honests.dup.concat(honests_by_n).uniq,
                unkinds.dup.concat(unkinds_by_n).uniq,
                next_unkown
            )
            ret = r if r > ret
        end
    end
    return ret
end

puts calc([],unkinds,all)