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 Lua (Lua 5.3.5) (judge verdict: Accepted)

local n=io.read("n")
local testimony={}
for i=0,n-1 do
    testimony[i]={}
    local a=io.read("n")
    for j=1,a do
        local x,y=io.read("n","n")
        table.insert(testimony[i],{x-1,y})
    end
end

local function check(bit)
    local checker=true
    for i=0,n-1 do
        if (bit & 1<<i)>0 then
            for k,v in pairs(testimony[i]) do
                local x=v[1]
                local y=v[2]
                if y==1 and (bit & 1<<x)==0 then
                    checker=false
                end
                if y==0 and (bit & 1<<x)>0 then
                    checker=false
                end
            end
        end
    end
    return checker
end

local max=0
for bit=0,(1<<n)-1 do
    if check(bit) then
        local counter=0
        for i=0,n-1 do
            if (bit & (1<<i))>0 then
                counter=counter+1
            end
        end
        max=math.max(counter,max)
    end
end
print(max)