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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApp21 {
    class Program {
        static void Main(string[] args) {
            //入力シリーズ
            int N = int.Parse(Console.ReadLine());

            int[] A = new int[N];

            int[][] x = new int[N][];
            int[][] y = new int[N][];

            for (int i = 0; i < N; i++) {
                A[i] = int.Parse(Console.ReadLine());

                x[i] = new int[A[i]];
                y[i] = new int[A[i]];

                for (int j = 0; j < A[i]; j++) {
                    var p = Console.ReadLine().Split(' ').Select(q => int.Parse(q)).ToArray();
                    x[i][j] = p[0] - 1;
                    y[i][j] = p[1];
                }
            }

            //1がhonest, 0がdishonestである

            //解答
            int ans = 0;

            //N人のhonest or dishonestについて、ビットで全可能性を探索
            for (int s = 0; s < (1 << N); s++) {
                //全員のパターンのstringを作成する
                string st = "";
                for (int i = 0; i < N; i++) {
                    if ((s & (1 << (N - i - 1))) != 0) {
                        st += "1";
                    } else {
                        st += "0";
                    }
                }

                //カウンタ
                int cnt = 0;

                //Console.WriteLine(st);

                //左からi番目の人のビットが立っている（=honest）ときに、妥当性を調べる
                for (int i = 0; i < N; i++) {
                    if ((s & (1 << (N - i - 1))) != 0) {
                        if (A[i] == 0) {
                            cnt++;
                            //Console.WriteLine("a");
                        }

                        for (int j = 0; j < A[i]; j++) {
                            if (st[x[i][j]].ToString() == y[i][j].ToString()) {
                                cnt++;
                                //Console.WriteLine("b");
                            } else if (x[i][0] == -1) {
                                cnt++;
                                //Console.WriteLine("d");
                            } else {
                                //ダメだったときはループを抜ける
                                goto skip;
                            }
                        }
                    }
                }

                ans = Math.Max(ans, cnt);

                skip:;
            }


            Console.WriteLine(ans);
        }
    }
}