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: Accepted)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(){
  int N;
  cin >> N;
  vector<int> A(N);
  vector<vector<int>> x(N);
  vector<vector<bitset<1>>> y(N);
  for (int i=0; i<N; i++) {
    cin >> A[i];
    for (int j=0; j<A[i]; j++) {
      int a;
      bitset<1> b;
      cin >> a >> b;
      x.at(i).emplace_back(a);
      y.at(i).emplace_back(b);
    }
  }
  bool flag = false;
  int count = 0;
  for (int tmp=0; tmp < (1 << N); tmp++) {
    flag = true;
    bitset<15> katei(tmp);
    for (int i=0; i<N; i++) {
      if (!katei[i]) continue;
      for (int j=0; j<A[i]; j++) {
        if ((y.at(i).at(j))!=(katei.test(x.at(i).at(j)-1))) flag = false;
      }
    }
    if (flag) {
      int tmp_count = 0;
      for (int i=0; i<N; i++) {
        if (katei[i]) tmp_count++;
      }
      if (tmp_count > count) count = tmp_count;
    }
  }
  cout << count << endl;
}