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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ

#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))

int main(void){
    int N;
    scanf("%d",&N);

    //Map[i][j]...iの人がjの人に対してなんて言及しているか(1→正直,0→不親切,-1→言及なし)
    int Map[N][N];
    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++) Map[i][j]=-1;
    }

    //Mapを埋める
    for(int i=0;i<N;i++){
        int A;
        scanf("%d",&A);
        int x,y;
        for(int j=0;j<A;j++){
            scanf("%d%d",&x,&y);
            Map[i][x-1]=y;
        }
    }

    int ans=0;
    for(int i=0;i<(1<<N);i++){
        int flg=1;
        for(int j=0;j<N;j++){
            if(!flg) break;

            int person; //jが正直ものなら1,真偽不明なら0
            if(i&(1<<j)) person=1;
            else person=0;

            for(int k=0;k<N;k++){
                if((i&(1<<k))==0) continue;
                
                if(Map[k][j]!=person){
                    if(Map[k][j]!=-1){
                        flg=0;
                        break;
                    }
                }
            }
        }
        if(flg) ans=MAX(ans,__builtin_popcount(i));
    }
    printf("%d\n",ans);

    return 0;
}