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 (1.15.1) (judge verdict: Accepted)

// 標準入力から一行を読み取り、指定の型に変換する
fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

// 標準入力から一行を読み取り、空白文字で分割し、各要素を指定の型に変換する関数
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace()
        .map(|e| e.parse().ok().unwrap()).collect()
}

use std::cmp::*;

fn main() {
    let n: usize = read();
    // 証言の配列
    let mut testimonies= vec![vec![]; n];
    for index in 0..n
    {
        // 各人の証言を入力
        let a_n: usize = read();
        for _ in 0..a_n
        {
            let testimony: Vec<u32> = read_vec();
            testimonies[index].push((testimony[0] - 1, testimony[1]));
        }
    }

    // 正直者の最大人数
    let mut honest_count_max: u32 = 0;
    // 証言の検証 bit全探索
    'outer: for bp in 0..1 << n
    {
        // 正直者である人物の証言を調べ、矛盾を確かめる.
        for index in 0..n
        {
            if bp >> index & 1 == 1
            {
                for t_index in 0..testimonies[index].len()
                {
                    // 矛盾の発見
                    if bp >> testimonies[index][t_index].0 & 1 != testimonies[index][t_index].1
                    {
                        continue 'outer;
                    }
                }
            }
        }

        // count_ones()はビット列中の'1'の数を返す
        honest_count_max = max(honest_count_max, bp.count_ones());
    }

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