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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;

fn main() {
  input! {
    n: usize,
  }
  let mut patterns:Vec<Vec<usize>> = vec![vec![10;n];n];
  for i in 0..n {
    input! {
      p: usize,
    }
    
    for _ in 0..p {
      input! {
        a: usize,
        b: usize
      }
      patterns[i][a-1] = b;
    }
  }

  let max = 2usize.pow(n as u32);
  let mut result = 0;
  for i in 1..=max {
    let mut memo = vec![0;n];
    let mut count = 0;
    for index in 0..n {
      if i>>index & 1 == 1 {
        memo[index] = 1;
        count += 1;
      }
    }
    
    let mut success = true;
    for index in 0..n {
      if memo[index] != 1 {
        continue
      }

      for (i, v) in patterns[index].iter().enumerate() {
        if *v == 10 { continue }
        if memo[i] != *v {
          success = false;
          break;
        }
      }
      if !success {
        break
      }
    }

    if success {
      result = std::cmp::max(result, count);
    }
  }
  println!("{}", result);
}