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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.*

fun main(args: Array<String>) {
    val sc = Scanner(System.`in`)
    val n = sc.nextInt()
    val a = (0 until n).map {
        (0 until sc.next().toInt()).map {
            sc.next().toInt() - 1 to sc.next().toInt()
        }
    }
    println(problem147c(n, a))
}

fun problem147c(
    n: Int,
    a: List<List<Pair<Int, Int>>>
): Int {
    fun judge(bit: Int): Boolean {
        for (i in 0 until n) {
            if (bit and (1 shl i) == 0) continue
            for (xy in a[i]) {
                val (x, y) = xy
                if (y == 1 && (bit and 1 shl x == 0)) return false
                if (y == 0 && (bit and 1 shl x != 0)) return false
            }
        }
        return true
    }

    var res = 0
    var bit = 0
    while(bit < (1 shl n)) {
        if (judge(bit)) {
            var count = 0
            for (i in 0 until n) {
                if (bit and (1 shl i) != 0) count++
            }
            res = Math.max(res, count)
        }
        bit++
    }
    return res
}