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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;

public class Main {

	public static void main(String[] args) throws Exception {
		// TODO 自動生成されたメソッド・スタブ
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		String W = in.readLine();
		int N = Integer.parseInt(W)+1;
		int[][] A = new int[N][N];
		for(int i=1;i<N;i++){
			Arrays.fill(A[i], -1);
		}
		for(int i=1;i<N;i++){
			W = in.readLine();
			int t = Integer.parseInt(W);
			for(int k=0;k<t;k++){
				W = in.readLine();
				String[] Ws = W.split(" ");
				A[i][Integer.parseInt(Ws[0])]= Integer.parseInt(Ws[1]);
			}
		}
		int ans = 0;
		for(int i=(int)Math.pow(2, N-1)-1;i>0;i--){

			String truth ="0"+String.format("%0"+(N-1)+"d",Integer.parseInt(Integer.toBinaryString(i)));
			char[] tmp = truth.toCharArray();
			int[] t = new int[tmp.length];

			for(int ti = 1;ti<t.length;ti++){
				t[ti] = (char)(tmp[ti]-'0');
			}
			boolean unmachFlg = false;
			for(int x=1;x<N;x++){
				for(int y=1;y<N;y++){
					if(t[y] == 1 && A[x][y] !=-1 && A[x][y] != t[x]){
						unmachFlg = true;
						break;
					}
				}
				if(unmachFlg){
					break;
				}
			}
			if(!unmachFlg){
				ans = Math.max(ans, Integer.bitCount(i));
			}
		}

		System.out.println(ans);


	}


}