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>
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define rep(i,n) for (int i = (0); i < (n); ++i)
#define rng(x) (x).begin(),(x).end()
#define rrng(x) (x).rbegin(),(x).rend()
#define debug(x) cout << #x << " = " << (x) << endl;
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define uni(x) x.erase(unique(rng(x)),x.end())
#define SP << " " <<
using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using pii = pair<int, int>;
using tiii = tuple<int, int, int>;
const int mod = 1000000007;
const double EPS = 1e-9;
const int INF = 1<<30;
const ll INFLL = 1LL<<60;
//cout << fixed << setprecision(10);

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

int n;
int a[20];
int x[20][20];
int y[20][20];

int counter(int x) {
  if (x == 0) return 0;
  return counter(x >> 1) + (x&1);
}

int main () {
  cin >> n;
  rep(i,n) {
    cin >> a[i];
    rep(j,a[i]) {
      cin >> x[i][j] >> y[i][j];
      x[i][j]--;
    }
  }

  int ans = 0;
  rep(b,1<<n) {
    bool ok = true;
    rep(i,n) {
      if (!(b&(1<<i))) continue;
      // 正直者の証言が矛盾しないか確認
      rep(j,a[i]) {
        //if ((b&(1<<x[i][j])) ^ y[i][j]) ok = false;
        //b&(1<<x[i][j])は1より大きな値になりうり、xorの計算が狂うのでダメ
        //1^1 -> 0, 2^1 -> 3

        if (((b>>x[i][j])&1) ^ y[i][j]) ok = false;

        //printf("b = %d, b&(1<<x[i][j]) = %d, y[i][j] = %d\n", b, b&(1<<x[i][j]), y[i][j]);
        //printf("b = %d, (b>>x[i][j])&1 = %d, y[i][j] = %d\n", b, (b>>x[i][j])&1, y[i][j]);
      }
    }
    if (ok) ans = max(ans, counter(b));
    //if (ok) debug(b);
  }
  cout << ans << endl;
  return 0;
}