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)

#![allow(non_snake_case)]
#![allow(unused_imports)]
use proconio::input;
use std::collections::HashMap;
use std::collections::HashSet;
use std::cmp::max;
// use std::cmp::min;


fn main() {
    input! {
        N: usize,
    }

    let mut XYs: Vec<Vec<(usize, i64)>> = vec![Vec::new(); N];

    for n in 0..N {
        input! {
            A: usize,
            mut tmpXYs: [(usize, i64); A],
        }

        // 0-origin
        for tpl_ref in tmpXYs.iter_mut() {
            tpl_ref.0 -= 1;
        }

        XYs[n].extend_from_slice(&tmpXYs);
    }



    // N<=15なので、2^15のビット全探索でいけそう
    // 正直者である人たちの集合を全探索
    // https://qiita.com/Cassin01/items/246e795d0d9003c64d0f
    let mut ans = 0;
    for i in 0..1 << N {
        let mut cand = 0;

        // 正直者によって得られた証言セット
        let mut map: HashMap<usize, i64> = HashMap::new();

        let mut is_ok = true;
        let mut honest_set: HashSet<usize> = HashSet::new();

        // 証言記録フェーズ
        for j in 0..N {
            if (1 << j) & i == 0 {
                cand += 1;
                honest_set.insert(j);

                for &(x, y) in XYs[j].iter() {
                    match map.get(&x) {
                        Some(&val) => {
                            if val != y {
                                is_ok = false;
                                break;
                            }
                        }
                        None => {
                            map.insert(x, y);
                        }
                    }
                }
            }
        }

        if !is_ok {
            eprintln!("NG ptn1: {:?}", honest_set);
            continue;
        }

        // 証言の中で言及された正直者/不親切が実際にそうなのか確認フェーズ
        for j in 0..N {
            let j_is_honest = (1 << j) & i == 0;
            let honest_val = if j_is_honest { 1 } else { 0 };

            match map.get(&j) {
                Some(&val) => {
                    if val == honest_val {
                        // ok
                    } else {
                        is_ok = false;
                    }
                }
                None => {
                    // ok
                }
            }
        }

        if is_ok {
            ans = max(ans, cand);
        } else {
            eprintln!("NG ptn2: {:?}", honest_set);
        }
    }

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