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)

fun main(args: Array<String>) {
    abc147c()
}

private fun bits(x: Int): Int {
    var ret = 1
    repeat(x) { ret *= 2 }
    return ret - 1
}

fun abc147c() {
    val n = readLine()!!.toInt()
    val aList = mutableListOf<Int>()
    val xyList = mutableListOf<List<Pair<Int, Int>>>()
    (0 until n).forEach {
        aList.add(readLine()!!.toInt())
        xyList.add((1..aList[it]).map { readLine()!!.split(' ').map { it.toInt() } }.map { Pair(it[0] - 1, it[1]) })
    }

    var answer = 0
    for (i in 0 until bits(n)) {
        val list = (1..n).map { -1 }.toMutableList()
        val all = xyList.withIndex().all { iv ->
            iv.value.all {
                val (target, value) = it
                val honest = (i shr iv.index) % 2 == 0
                val value1 = if (honest) value else (if (value == 0) 1 else 0)
                if (list[iv.index] != -1 && list[iv.index] != if (honest) 1 else 0) {
                    false
                } else {
                    if (list[target] == -1) {
                        list[target] = value1
                        true
                    } else {
                        list[target] == value1
                    }
                }
            }
        }
        if (all) {
            answer = Math.max(answer, list.count { it == 1 })
        }
    }

    println(answer)
}