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, a, b) for (int i = (a); i < (int)(b); i++)
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(a) (a).begin(), (a).end()
#define allg(a) (a).begin(), (a).end(), greater<int>()

using vi = vector<int>;
using vvi = vector<vi>;
using vs = vector<string>;
using pii = pair<int, int>;

const int INF = 1001001001;
const int MOD = 1000000007;
const long long INFL = (1LL << 60);
const double EPS = 1e-9;

int N;
vector<vector<pii>> v;  // pii: (x,y)

bool judge(int bit) {
  rep(i, N) {
    // person i
    if (!(bit & (1 << i))) continue;

    for (auto xy : v[i]) {
      int x = xy.first;
      int y = xy.second;
      if (y == 1 && !(bit & (1 << x))) return false;  // contradiction
      if (y == 0 && (bit & (1 << x))) return false;   // contradiction
    }
  }
  return true;
}

int main() {
  cin >> N;
  v.resize(N);
  rep(i, N) {
    int A;
    cin >> A;
    v[i].resize(A);
    rep(j, A) {
      cin >> v[i][j].first >> v[i][j].second;
      v[i][j].first--;
    }
  }

  int res = 0;
  for (int bit = 0; bit < (1 << N); bit++) {
    if (judge(bit)) {
      int cnt = 0;
      rep(i, N) {
        if (bit & (1 << i)) cnt++;
      }
      res = max(res, cnt);
    }
  }
  cout << res << endl;
}