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: Wrong Answer)

/*...................bismillahir rahmanir rahim.....................*/
#include<bits/stdc++.h>

#define m_p make_pair
#define PB push_back
#define FF first
#define SS second
#define PII pair<int,int>
#define PSS pair<string,string>
#define MX 1000007
#define MOD 1000000007
#define SI(a)({ ;scanf("%d",&a);a;})
#define SC(a)({ ;scanf("%c",&a);a;})
#define SL(a)({ ;scanf("%lld",&a);a;})
#define SD(a)({ ;scanf("%lf",&a);a;})
#define FI freopen("input.txt","r+",stdin);
#define FO freopen("output.txt","w+",stdout);
using namespace std;

typedef long long LL; 

/*............start.............*/
int Set(int n,int pos){ return n=n | (1<<pos); }
int Reset(int n,int pos){ return n=n & ~(1<<pos); }
bool check(int n,int pos){ return (bool)(n & (1<<pos)); }
int N;
int tes[16][16];
// int DP[1<<15];
int call(int mask)
{
    // if(DP[mask]!=-1) return DP[mask];
    bool t=true;
    for(int i=1;i<=N;i++)
    {
        if(check(mask,i)==1)
        for(int j=1;j<=N;j++)
        {
           if(i!=j && tes[i][j]!=-1) t&=~(check(mask,j)^tes[i][j]);
        }
    }
    if(t)
    {
        int r=0;
        for(int i=1;i<=N;i++) if(check(mask,i)) r++;
        return r; 
    }
    return 0;
}
void solve()
{
    // int n;
    cin>>N;
    memset(tes,-1,sizeof(tes));
    // memset(DP,-1,sizeof(DP));
    for(int i=1;i<=N;i++)
    {
        int x,y,t;
        cin>>x;
        while(x--)
        {
            cin>>y>>t;
            tes[i][y]=t;
        }
    }
    int res=0;
    for(int i=0;i<(1<<15);i++)
    {
        res=max(res,call(i));
    }
    cout<<res<<endl;
}

int main()
{
    ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
    int q=1;
    // cin>>q;
    for(int kase=1;kase<=q;kase++)
    {
        //cout<<"Case "<<kase<<": ";
        solve();
    }
    return 0;
}