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 MOD 1000000007
#include <map>
#define REP(i, n) for(int i = 0; i < n; i++)
#define REPR(i, n) for(int i = n; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define INF 2e9
#define ALL(v) v.begin(), v.end()
typedef long long ll;

int main(){
    ll n;
    cin >> n;
    vector<ll> A(n);
    vector<vector<ll>> X(n,vector<ll>(15));
    vector<vector<ll>> Y(n,vector<ll>(15));
    REP(i,n){
    cin >> A[i];
        REP(j,A[i]){
            cin >> X[i][j] >> Y[i][j];
        }
    }
    
    ll mat[2][15];
    ll ans = 0 ;
    ll b = 1;
    REP(i,n){
        b *= 2;
    }
    REP(i,2){
        REP(j,15){
            mat[i][j] = 0;
        }
    }
    REP(i,b){
        ll mat[2][15];
        REP(i,2){
            REP(j,15){
                mat[i][j] = 0;
            }
        }
        ll B[15];
        ll j = i;
        REP(p,15){
            B[p] = j%2;
            j /= 2;
        }
        REP(q,n){
            if(B[q] == 1){
                mat[0][q] += 1;
                REP(k,A[q]){
                    if(Y[q][k] == 1){
                        mat[0][X[q][k]-1] += 1;
                    }
                    else{
                        mat[1][X[q][k]-1] += 1;
                    }
                }
            }
            if(B[q] == 0){
                mat[1][q] += 1;
            }
        }
        /*
        REP(i,15){
            cout << B[i];
        }
        cout << endl;
        REP(i,2){
            REP(j,15){
                cout << mat[i][j];
            }
            cout << endl;
        }
        */
        ll memo = 0;
        REP(r,15){
            memo += B[r];
        }
        ll memo2 = 0;
        REP(r,15){
            memo2 += mat[0][r] * mat[1][r];
        }
        
        if(memo2 == 0){
            if(ans < memo){
                ans = memo;
            }
        }
    }
    cout << ans << endl;
}