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: Wrong Answer)

use std::io::prelude::*;
use std::io::stdin;
use std::str::FromStr;

fn main() {
    let n: usize = read();
    let mut v = vec![];

    for _ in 0..n {
        let a: usize = read();
        let xy: Vec<(usize, usize)> = (0..a).map(|_| (read(), read())).collect();
        v.push(xy);
    }

    let mut v1 = vec![1; n];
    for (i, xy) in v.iter().enumerate() {
        if v1[i] == 1 {
            for &(x, y) in xy {
                if v1[x - 1] != y {
                    v1[i] = 0;
                }
            }
        }
    }

    for (i, xy) in v.iter().rev().enumerate() {
        if v1[i] == 1 {
            for &(x, y) in xy {
                if v1[x - 1] != y {
                    v1[i] = 0;
                }
            }
        }
    }
    println!("{}", v1.iter().filter(|x| **x == 1).count());
}

#[allow(dead_code)]
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|b| b.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}