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 Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

{-# LANGUAGE BangPatterns #-}
import Control.Monad
import Data.Maybe
import Debug.Trace
import Data.List
import qualified Data.Map.Strict as Map
import qualified Data.IntMap.Strict as IntMap
import qualified Data.Set as Set
import qualified Data.IntSet as IntSet
import Data.Functor
import Data.Array
-- import Data.Array.Unboxed
import Control.Monad.ST
import Data.Array.ST
import Data.Char
import Data.Bits

main :: IO ()
main = do
  n <- readLn :: IO Int
  xys <- replicateM n readxy :: IO [[[Int]]]
  print $ solve n xys

solve :: Int -> [[[Int]]] -> Int
solve n xys = if null ans then 0 else (maximum ans)
  where
    ans = map sum . filter isPossible $ candidate n
    isPossible candidate = go candidate xys
      where
        go :: [Int] -> [[[Int]]] -> Bool
        go [] _ = True
        go _ [] = True
        go (c:cs) (xy:xys)
          | isOK = go cs xys
          | otherwise = False
          where
            isOK = foldl' g True xy
            g acc (x:y:_) = acc && (f x y)
            f x y = if c == 1 
              then candidate!!(x-1) == y
              else candidate!!(x-1) /= y
              -- (c == 1) `xor` (candidate!!(x-1) /= y)

readxy = do
  a <- readLn :: IO Int
  xy <- replicateM a $ map read . words <$> getLine :: IO [[Int]]
  return xy


candidate n = replicateM n [0, 1]