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,j,N) for(int i = (j);i < (N);i++)
#define _rep(i,j,N) for(int i = (j);i <= (N);i++)
#define _dec(i,N,j) for(int i = (N-1);i >= (j);i--)
#define _rec(i,N,j) for(int i = (N);i >=(j);i--)
#define ALL(x) x.begin(),x.end()
#define MEM(a,n) memset(a,n,sizeof(a))
#define fst first
#define scd second
#define PB push_back
#define MP make_pair

typedef long long LL;
typedef long double LD;
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

int get_int(){int x;scanf("%d",&x);return x;}
LL get_ll(){LL x;scanf("%lld",&x);return x;}
double get_db(){double x;scanf("%lf",&x);return x;}

template<typename T> int chmin(T &a,T b){if(a > b){a = b;return 1;}else return 0;}
template<typename T> int chmax(T &a,T b){if(a < b){a = b;return 1;}else return 0;}
template<typename T>
ostream& operator<<(ostream& os,const vector<T>& v) {_for(i,0,v.size()) os << v[i] << " ";return os;}
template<typename T>
ostream& operator<<(ostream& os,const set<T>& v){for(auto a:v) os << a <<" ";return os;}
template<typename T1,typename T2>
ostream& operator<<(ostream& os,const pair<T1,T2>& v){os << v.fst <<" "<<v.scd << " ";return os;}
template<typename T1,typename T2>
ostream& operator<<(ostream& os,const map<T1,T2>& v){for(auto a:v) cout << a << endl;return os;}

static const int dx[8] = { -1,  0, 1, 0,  1, -1,  1, -1};
static const int dy[8] = {0,  1, 0,  -1,  1, -1, -1,  1};
const double pi  = acos(-1.0);
const double eps = 1e-8;
//const LL  INF  = 0x3f3f3f3f3f3f3f3f;
const int inf  = 0x3f3f3f3f;

const int maxn = 1e5+5;

int N;
vector<pii> G[20];

int is_ho(int s,int n){
    return 1 & (s>>(n-1));
}

int judge(int s){
    int OK = 1;
    _rep(i,1,N){
        if(is_ho(s,i)){
            for(pii b : G[i]){
                int jj = b.first;
                int yy = b.second;
                if(yy){
                    if(!is_ho(s,jj)) OK = 0;

                }else{
                    if(is_ho(s,jj)) OK = 0;
                }

            }
        }
    }

    return OK;
}

int ct(int s){
    int ret = 0;
    _for(i,0,N){
        if((s>>i)&1) ret++;
    }

    return ret;

}

int main()
{
    cin >> N;
    _rep(i,1,N){
        int A;
        cin >> A;
        _for(j,0,A){
            int x,y;
            cin >> x >> y;
            G[i].PB(MP(x,y));
        }
    }



    int full = (1<<N)-1;
    int ans = 0;
    _rep(i,0,full){
        if(judge(i))
            ans = max(ans,ct(i));
    }
    cout << ans << endl;



    return 0;
}