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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

n = int(input())
xy = []

for i in range(n):
    a = int(input())
    temp = {}
    for j in range(a):
        x, y = map(int, input().split())
        #temp[j] = [x, y]
        #temp.append({x: y})
        temp[x] = y
    xy.append(temp)

def solve(idx, isHonest, hList, cnt):
    if idx == n:
        return cnt
    infoMap = xy[idx]

    if isHonest:
        cnt += 1
        for (key, value) in infoMap.items():
            if hList[key] is None:
                hList[key] = value
            elif hList[key] != value:
                return 0
    else:
        for (key, value) in infoMap.items():
            if hList[key] is None:
                hList[key] = abs(value - 1)
            elif hList[key] != abs(value - 1):
                return 0

    return max(solve(idx+1, isHonest, hList, cnt), solve(idx+1, not isHonest, hList, cnt))


hList1 = [None] * (n+1)
hList1[1] = 1
hList2 = [None] * (n+1)
hList2[1] = 0
print(max(solve(0, True, hList1, 0), solve(0, False, hList2, 0)))