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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;

//template
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(a);i>(b);i--)
#define ALL(v) (v).begin(),(v).end()
typedef long long int ll; const int inf = 0x3fffffff; const ll INF = 0x3fffffffffffffff;
template<class T> inline bool chmax(T& a,T b){ if(a<b){a=b;return 1;}return 0; }
template<class T> inline bool chmin(T& a,T b){ if(a>b){a=b;return 1;}return 0; }
//template end

typedef pair<int,int> P;

int main(){
    int n; scanf("%d",&n);
    vector<int> a(n); vector<vector<P>> xy(n);
    rep(i,0,n){
        scanf("%d",&a[i]); xy[i].resize(a[i]);
        rep(j,0,a[i]){
            scanf("%d%d",&xy[i][j].first,&xy[i][j].second);
            xy[i][j].first--;
        }
    }
    int ans=0;
    rep(mask,0,1<<n){
        bool f=1;
        rep(i,0,n)if(mask>>i&1){
            rep(j,0,a[i]){
                if((mask>>i&1)==(mask>>xy[i][j].first&1)){
                    if(xy[i][j].second==0){
                        f=0; 
                        i+=n; j+=a[i]; break;
                    }
                }
                if((mask>>i&1)!=(mask>>xy[i][j].first&1)){
                    if(xy[i][j].second){
                        f=0; 
                        i+=n; j+=a[i]; break;
                    }
                }
            }
        }
        if(f)chmax(ans,__builtin_popcount(mask));
    }
    printf("%d\n",ans);
    return 0;
}