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: Wrong Answer)

#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>



using namespace std;
int main(void){

  int n;

  cin >> n;

  int a[n];
  int x[n][n];
  int y[n][n];
  int i,j,k,l;
  int flag=1;//1 true  0 false

  for(i=0;i<n;i++){
    cin >> a[i];
    for(j=0;j<a[i];j++){
      cin >> x[i][j];
      x[i][j]=x[i][j]-1;
      cin >> y[i][j];
    }
  }
  int ans=0;//this will be an answer
  //一人ずつそいつが正直者であると仮定して、矛盾があればつぎにいく
  for(i=0;i<n;i++){
    for(j=0;j<a[i];j++){
      for(k=0;k<n;k++){
        if(k!=i){
        for(l=0;l<a[k];l++){
          if(i==x[k][l]){
            if(y[i][j]==1&&y[k][l]==1){
              flag=1;
              //cout << "mondai2\n";
              //cout << i <<j<<k<<l;
            }
            else if(y[i][j]==1&&y[k][l]==0){
              flag=0;
              //cout << "mondai1\n";
              //cout << i <<j<<k<<l;
              break;
            }
            else if(y[i][j]==1&&y[k][l]==1){
              flag=0;
              //cout << "mondai1\n";
              //cout << i <<j<<k<<l;
              break;
            }

        }

          }
          if(flag==0){
            break;
          }
      }
      if(flag==0){
        break;
      }
    }
    }

    flag=1;


  }
  if(n==2){

  }
  cout << ans;


  return 0;
}