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)

#include <bits/stdc++.h>
using namespace std;
#define FOR(i, o, n) for (auto i = (o); (i) < (n); ++(i))
#define FOR1(i, n) FOR(i, 1, n+1)
#define REP(i, n) FOR(i, 0, n)
#define REPR(i, n) for (auto i = (n)-1; (i) >= 0; --i)
#define REPI(itr, v) for (auto itr = v.begin(); itr != v.end(); ++itr)
#define REPIR(itr, v) for (auto itr = v.rbegin(); itr != v.rend(); ++itr)
#define fi first
#define se second
#define ll long long
#define int long long
#define V2(name, h, w) vector<vector<int> > name((h), vector<int>(w));

#define dl(line) cerr << "RUN: until " << (line) << "th line." << endl
#define dv(name, value) cerr << (name) << " is " << (value) << endl
#define da(v)  cerr << "["; REPI(itr, v){cerr << *itr << (itr != --v.end() ? ", " : "");} cerr << "]" << endl


struct Info {
  int x = 0; 
  bool y = false;
};

signed main(){
  int N; cin >> N;
  vector<vector<Info> > v(0);
  REP(x, N) {
    int A; cin >> A;
    vector<Info> inpV(A);
    REP(y, A) {
      cin >> inpV.at(y).x >> inpV.at(y).y;
    }
    v.push_back(inpV);
  }

  vector<bool> flags(N+1);
  int maxSum = 0;

  REP(bit, (1<<N)) {
    REP(x, N) {
      if(bit & (1<<x)){
        flags.at(x+1) = true;
      } else {
        flags.at(x+1) = false;
      }
    }

    bool isValid = true;
    // dv("bit", bit);
    FOR1(flagX, N) {
      if(flags.at(flagX)) {
        REP(y, (ll)v.at(flagX-1).size()) {
          if(v.at(flagX-1).at(y).y != flags.at((v.at(flagX-1).at(y).x))) {
            isValid = false;
          }
        }
      }
    }
    if (isValid) {
      // da(flags);
      int sum = 0;
      FOR1(x, N) {
        sum += flags.at(x) ? 1 : 0;
      }
      maxSum = max(maxSum, sum);
    }
  }

  cout << maxSum << endl;
}