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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <iostream>
#include <math.h>

using namespace std;


//引数numを2進数に直して配列true_personに左から格納する
void binary_num(int num, int N, int true_person[]){
  int i=1;
  while(num != 0){
    true_person[i] = num%2;
    i++;
    num /= 2;
  }
}


//true_person配列を参照することで，正直者の人数を返す関数
int honest_count(int true_person[], int N){
  int count = 0;
  for(int i=1; i<=N; i++)
    if(true_person[i] == 1)
      count++;
  return count;
}


int check(int x[15][15], int y[15][15], int A[],
		 int N, int true_person[]){
  int i, j;
  for(i=1; i<=N; i++){
    if(true_person[i] == 1){
      for(j=1; j<=A[i]; j++){
	if(y[i][j] != true_person[ x[i][j] ])
	  return -1;
      }
    }
  }
  return honest_count(true_person, N);
}


int main(void){
  int N, i, j;
  cin >> N;
  int A[N];
  int x[15][15], y[15][15];
  for(i=1; i<=N; i++){ //入力
    cin >> A[i];
    for(j=1; j<=A[i]; j++)
      cin >> x[i][j] >> y[i][j];
  }
    
  //人iが正しいかどうか(正しいなら1，そうでないなら0)
  int true_person[15];
  
  /*binary_numにより，人iが正しいかどうかをtrue_personに格納する．
    例えば，i=20回目のループを考える．
    20は2進数で00101(左から数える)であり，これは
    人1が嘘つき，人2が嘘つき，人3は正直，人4は嘘つき，人5は正直
    であることを示しており，それをi=0からi=2^N-1まで繰り返すことにより
    全パターンを探索することができる．
   */
  int count_max = 0;
  int count;
  for(i=0; i<pow(2, N); i++){
    binary_num(i, N, true_person);    
    count = check(x, y, A, N, true_person);
    if(count_max < count)
      count_max = count;
  }
  cout << count_max << endl;
  return 0;
}