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};
use std::cmp::max;
#[allow(unused_mut)]
#[fastout]
fn main() {
    input! {
        n: usize
    }
    let mut info = vec![vec![]; n];
    for i in 0..n {
        input! {
            a: usize
        }
        for _ in 0..a {
            input! {
                p: usize,
                tf: usize
            }
            info[i].push((p - 1, tf));
        }
    }

    let mut res = 0;
    for bit in 0usize..(1 << n) {
        let mut can = true;
        for i in 0..n {
            //仮定した正直者だけ発言に矛盾がないか吟味する
            if bit & (1 << i) != 0 {
                for (p, tf) in info[i].iter() {
                    //もし矛盾していたらその組み合わせは没
                    if !((*tf == 0 && (bit & (1 << *p)) == 0)
                        || (*tf == 1 && (bit & (1 << *p)) != 0))
                    {
                        can = false;
                    }
                }
            }
        }

        if can {
            res = max(res, bit.count_ones());
        }
    }

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