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;
typedef long long ll;
#define mp make_pair
#define pb push_back
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)


int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);

  int n;
  cin >> n;
  int a[n];
  vector<pair<int, int> > dic[n];
  rep(i,n){
    cin >> a[i];
    rep(j,a[i]){
      int x,y;
      cin >> x >> y;
      --x;
      dic[i].pb(mp(x,y));
    }
  }
  int ans=0;
  rep(k,(1<<n)){
    bool ok=true;
    rep(i,n)rep(j,a[i]){
      if(k & (1<<i)){
	int x = dic[i][j].first;
	int y = dic[i][j].second;
	//	if(k==3){
	//	  cout << bitset<3>( k && (1<<i)) << endl;
	//	  cout << "k, x " << bitset<3>(k)<< " " << bitset<3>((1<<x)) << endl;
	//      cout << x << y << endl;
	//	}
	if(y==1){//xは親切
	  if((k & (1<<x))){}
	  else ok=false;
	}
	else{//xは不親切
	  if((k & (1<<x))) ok=false;
	}
      }
    }
    //    if(k==3) cout << ok << endl;
    //    cout << ok << endl;
    if(ok) ans = max(ans, __builtin_popcount(k));
  }
  cout << ans << endl;
  
  
  return 0;
    

}