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: Runtime Error)

import java.util.*;

public class Main {
	public static void main(String[] args)  {
		Scanner scan = new Scanner(System.in);
    	
		int n=Integer.parseInt(scan.next());
		int a,x,y;
		int[][] xy=new int [15][15];//誰が　誰を　1or0
		for (int i=0;i<15;++i) {
			Arrays.fill(xy[i], -1);			
		}

		for (int i=0;i<n;++i) {
			a=Integer.parseInt(scan.next());
			for (int k=0;k<a;++k) {
				x=Integer.parseInt(scan.next());
				y=Integer.parseInt(scan.next());
				xy[a][x]=y;
			}
		}
		
		// solve
		int max=0;
		int[] d=new int[15];
		Arrays.fill(d, -1);
		for (int i=0;i<1<<n;++i) {//2のn情の組み合わせ
			for (int k=0 ;k<n;++k) {
				if ((i>>k&1)>0) {//k番目の人が正直もの		
					d[k]=1;
				}else {//k番目の人が不親切な人
					d[k]=0;
				}
			}
			
			//判定
			boolean bool =true;
			int count=0;//正直もののカウント
			for (int k=0 ;k<n;++k) {
					if (d[k]==1) {//ｋ番目の人が正直ものだったら
						++count;
						for (int j=0;j<n;++j) {//ｋ番目の人の証言を確認
							if (xy[k][j]==-1) continue;
							else if(xy[k][j]!=d[j]) bool=false;//証言が違う
						}

				}
			}if (bool) {
				max=Math.max(max, count);
				}
		}
        
       System.out.print(max);
        
        

}
}
//end