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::{fastout, input};

#[fastout]
fn main() {
    input! {n: usize};

    let mut v: Vec<Vec<i64>> = vec![vec![-1; 15]; 15];

    for i in 0..n {
        input! {a: usize};
        for _ in 0..a {
            input! {x: usize, y: i64};
            v[i][x - 1] = y;
        }
    }

    let mut ans = 0;
    for i in 0..1 << n {
        let mut d = vec![0; n];
        let mut ok = true;
        for j in 0..n {
            if i >> j & 1 == 1 {
                d[j] = 1;
            }
        }

        for j in 0..n {
            if d[j] == 1 {
                for k in 0..n {
                    if v[j][k] == -1 {
                        continue;
                    }

                    if v[j][k] != d[k] {
                        ok = false;
                        break;
                    }
                }
            }
        }
        if ok {
            ans = std::cmp::max(ans, (i as i32).count_ones());
        }
    }

    println!("{}", ans);
}