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)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) REP(i,0,n)
#define REP(i,s,e) for(int i=(s); i<(int)(e); i++)
#define repr(i, n) REPR(i, n, 0)
#define REPR(i, s, e) for(int i=(int)(s-1); i>=(int)(e); i--)
#define pb push_back
#define all(r) r.begin(),r.end()
#define rall(r) r.rbegin(),r.rend()
#define fi first
#define se second

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const ll INF = 1e9;
const ll MOD = 1e9 + 7;
double EPS = 1e-8;

template<typename T> T chmax(T& a, const T& b){return a = (a > b ? a : b);}
template<typename T> T chmin(T& a, const T& b){return a = (a < b ? a : b);}

int main(){
    int n;
    cin >> n;
    vector<int> a(n);
    vector<vector<int>> x(n), y(n);
    rep(i, n) {
        cin >> a[i];
        x[i].resize(a[i]);
        y[i].resize(a[i]);
        rep(j, a[i]) {
            cin >> x[i][j] >> y[i][j];
            x[i][j]--;
        }
    }
    int ans = 0;
    rep(mask, 1<<n) {
        bool f = true;
        rep(i, n) {
            rep(j, a[i]) {
                bool a = mask&(1<<i);
                bool b = mask&(1<<x[i][j]);
                bool c = y[i][j];
                f &= ((a^b)^c);
            }
        }
        if(f) chmax(ans, __builtin_popcount(mask));
    }
    cout << ans << endl;
    return 0;
}