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>

#ifndef M_PI
#define M_PI 3.14159265358979
#endif
#define deg_to_rad(deg) (((deg) / 360) * 2 * M_PI)
#define rad_to_deg(rad) (((rad) / 2 / M_PI) * 360)

using namespace std;

typedef long long ll;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;

const ll INF = 1e15;
const ll MOD = 1e9 + 7;

int main()
{
	ll i, j, k;
	ll N;
	cin >> N;
	vector<vll> X, Y;
	for (i = 0; i < N; i++)
	{
		ll a;
		cin >> a;
		vll xi, yi;
		for (j = 0; j < a; j++)
		{
			ll xij, yij;
			cin >> xij >> yij;
			xij--;
			xi.push_back(xij);
			yi.push_back(yij);
		}
		X.push_back(xi);
		Y.push_back(yi);
	}
	ll ans = 0;
	ll honest = 0;
	for (honest = 0; honest < (1 << N); honest++)
	{
		// cout << bitset<3>(honest) << endl;
		bool ok = true;
		for (i = 0; i < N; i++)
		{
			if (honest & 1 << i)
			{
				for (j = 0; j < X[i].size(); j++)
				{
					ll x = X[i][j], y = Y[i][j];
					if (y)
					{
						if (!(honest & 1 << x))
						{
							ok = false;
						}
					}
					else
					{
						if (honest & 1 << x)
						{
							ok = false;
						}
					}
				}
			}
		}
		if (!ok)
		{
			continue;
		}
		else
		{
			ans = max(ans, (ll)__builtin_popcountll(honest));
		}
	}
	cout << ans << endl;

	return 0;
}