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.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner sca = new Scanner(System.in);
		int n = sca.nextInt();
		int x[][] = new int[16][16];
		int y[][] = new int[15][15];
		int ai[] = new int[15];
		for(int i = 1; i < n+1; i++) {
			ai[i] = sca.nextInt();
			for(int j = 1; j < ai[i] + 1; j++) {
				x[i][j] = sca.nextInt();
				y[i][j] = sca.nextInt();
			}
		}

		// xi人目が正直なものと仮定したときでループさせ、
		// 矛盾が発生したとき、xi人目を嘘つきとして次のループへ
		int tr = 1;
		int result = 0;
		int big = 0;
		int human_katei[] = new int[n+1];

		while(tr <= n) {
			for(int i = 1; i < n+1; i++) {
				if(i == tr) {
					human_katei[i] = 1;
				}
				else {
					human_katei[i] = 3;
				}
			}
			for(int j = 1; j < ai[tr] + 1; j++) {
				human_katei[x[tr][j]] = y[tr][j];
			}
			int fal = 0;
			// tr人目が正直者と仮定する。
			for(int i = 1; i < n+1; i++) {
				if(i != tr) {
					int troo = 0;
					for(int j = 1; j < ai[i] + 1; j++) {
						if(human_katei[x[i][j]] == y[i][j] && human_katei[x[i][j]] != 3) {
							troo++;
						}
					}
					if(troo == ai[i]) {
						human_katei[i] = 1;
					}
					else {
						fal++;
					}
				}
			}
			if(fal == n-1) {
				for(int i = 1; i < n+1; i++) {
				human_katei[i] = 0;
				}
			}
			result = 0;
			for(int i = 1; i < n+1; i++) {
				if(human_katei[i] == 1) {
					result++;
				}
			}
			if(result > big) {
				big = result;
			}
			tr++;
		}

		System.out.println(big);
	}
}