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)

// https://atcoder.jp/contests/abc147/tasks/abc147_c
#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) FOR(i, 0, n)
#define REPR(i, n) for (int i = n - 1; i >= 0; i--)
#define FOR(i, s, n) for (int i = (s), i##_len = (n); i < i##_len; ++i)
#define ALL(obj) (obj).begin(), (obj).end()
#define ALLR(obj) (obj).rbegin(), (obj).rend()
#define DIV(a, b) ((a - 1) / b + 1)
const int SZ = 16;

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int N, ans = 0;
    cin >> N;
    vector<vector<int>> t(N), f(N);
    REP(i, N) {
        int A;
        cin >> A;
        REP(j, A) {
            int x, y;
            cin >> x >> y;
            if (y == 1) {
                t[i].emplace_back(x - 1);
            } else {
                f[i].emplace_back(x - 1);
            }
        }
    }
    REP(i, (1LL) << N) {
        bitset<SZ> bt, bf;
        // 矛盾したらダメ
        bool flag = true;
        REP(j, N) {
            if (i & (1LL << j)) {
                // trueリスト
                for (auto &x : t[j]) {
                    if (!(i & (1LL << x))) {
                        flag = false;
                        break;
                    }
                    bt.set(x);
                }
                // falseリスト
                for (auto &x : f[j]) {
                    bf.set(x);
                }
            }
        }
        REP(j, N) {
            if (bt[j] & bf[j]) {
                flag = false;
                break;
            }
        }
        if (flag) {
            ans = max(ans, __builtin_popcount(i));
        }
    }
    cout << ans << endl;
    return 0;
}