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 <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <iostream>
#include <istream>
#include <iterator>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#include <tuple>
#include <iomanip>
#include <climits>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
//typedef pair<ll, ll> P;
#define rep(i, n) for(ll i = 0; i < (n); i++)
#define revrep(i, n) for(ll i = (n)-1; i >= 0; i--)
#define pb push_back
#define f first
#define s second
#define chmin(x, y) x = min(x, y);
#define chmax(x, y) x = max(x, y);

//const ll INFL = LLONG_MAX;//10^18 = 2^60
const ll INFL = 1LL<<60;
//const int INF = INT_MAX;
const ll INF = 1 << 30;//10^9
ll MOD = 1000000007;
//ll MOD = 998244353;

vector<ll> dy = {0, 1, 0, -1, 1, 1, -1, -1, 0};
vector<ll> dx = {1, 0, -1, 0, 1, -1, 1, -1, 0};

void pres(double A, ll x = 10){cout << fixed << setprecision(x) << A << endl;}
void BinarySay(ll x, ll y = 60){rep(i, y) cout << (x>>(y-1-i) & 1); cout << endl;}
ll cnt_bit(ll x){return __builtin_popcountll(x);}

ll pow_long(ll x, ll k){
  ll res = 1;
  while(k > 0){
    if(k % 2) res *= x;
    x *= x; k /= 2;
  }
  return res;
}

ll pow_mod(ll x, ll k){
  ll res = 1;
  while(k > 0){
    if(k % 2){
      res *= x; res %= MOD;
    }
    x *= x; x %= MOD;
    k /= 2;
  }
  return res;
}

ll inverse(ll x){return pow_mod(x, MOD - 2);};

ll gcd(ll a, ll b){
    if(b == 0) return a;
    return gcd(b, a % b);
}

ll lcm(ll x, ll y){return x / gcd(x, y) * y;};


int A[16][16];

void solve(){
  int N;
  cin >> N;
  for(int i = 0; i < N; i++){
    int info;
    cin >> info;
    for(int j = 0; j < info; j++){
      int x, y;
      cin >> x >> y;
      x--;
      if(y == 1) A[i][x] = 1;
      else A[i][x] = -1;
    }
  }
  int ans = 0;
  for(int bit = 0; bit < 1<<N; bit++){
    bool ok = 1;
    for(int i = 0; i < N; i++){
      if(((bit>>i)&1) == 0) continue;
      for(int j = 0; j < N; j++){
        if(A[i][j] == 0) continue;
        if(A[i][j] == 1 && ((bit>>j)&1) == 0) ok = 0;
        if(A[i][j] == -1 && ((bit>>j)&1) == 1) ok = 0;
      }
    }
    if(ok) ans = max(ans, (int)cnt_bit(bit));
  }
  cout << ans << endl;
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  solve();
}