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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import java.util.stream.LongStream;

public class Main implements Runnable {

    private static int MOD = 1_000_000_007;

    public static void main(String[] args) {
        // Run with 32MB stack
        Thread thread = new Thread(null, new Main(), "", 32 * 1024 * 1024);
        thread.start();
    }

    @Override
    public void run() {
        final Scanner scanner = new Scanner(System.in);
        solve(scanner);
    }

    static void solve(Scanner scanner) {
        int N = Integer.parseInt(scanner.nextLine());
        List<List<List<Integer>>> n = new ArrayList<>();
        for (int i = 0; i < N; i++) {
            int A = Integer.parseInt(scanner.nextLine());
            List<List<Integer>> list = new ArrayList<>();
            for (int j = 0; j < A; j++) {
                int[] a = Arrays.stream(scanner.nextLine().split(" ")).mapToInt(Integer::parseInt).toArray();
                List<Integer> l = new ArrayList<>();
                l.add(a[0] - 1);
                l.add(a[1]);
                list.add(l);
            }
            n.add(list);
        }

        int ans = 0;
        loop:
        for (int i = 0; i < (1 << N); i++) {
            for (int j = 0; j < N; j++) {
                boolean t = ((1 << j) & i) > 0;
                if (t) {
                    for (List<Integer> l: n.get(j)) {
                        if (l.get(1) == 1) {
                            if (((1 << j) & l.get(0)) == 0) {
                                continue loop;
                            }
                        } else {
                            if (((1 << j) & l.get(0)) > 0) {
                                continue loop;
                            }
                        }
                    }
                } else {
                    for (List<Integer> l: n.get(j)) {
                        if (l.get(1) == 1) {
                            if (((1 << j) & l.get(0)) > 0) {
                                continue loop;
                            }
                        } else {
                            if (((1 << j) & l.get(0)) == 0) {
                                continue loop;
                            }
                        }
                    }
                }
            }
            ans = Math.max(ans, Integer.bitCount(i));
        }

        System.out.println(ans);
    }

}