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 Python (3.4.3) (judge verdict: Wrong Answer)

N = int(input())
remark = [0] * N
NGlist = [0] * N
grouplist = [0]*N
ans = 0
for i in range(N):
    remark[i] = [-1] * N
    NGlist[i] = [0] * N
    grouplist[i] = [0] * N


for k in range(N):
    n = int(input())
    for j in range(n):
        x, y = map(int, input().split())
        remark[k][x-1] = y
        if y == 0:
            NGlist[k][x-1] = 1
            NGlist[x - 1][k] = 1
        else:
            grouplist[k][x-1] = 1

for num in range(2 ** N):
    # print("num=")
    # print(num)
    group = []
    for h in range(N):
        if h not in group and num % 2 == 1:
            group.append(h)
            for s in range(h+1, N):
                if grouplist[h][s] == 1 and s not in group:
                    group.append(s)
            num //= 2
    len_group = len(group)
    # print(group)
    if ans >= len_group:
        flag = 1
    flag = 0

    for m in range(len_group):
        for n in range(m+1, len_group):
            if NGlist[m][n] == 1:
                flag = 1
                break
        if flag == 1:
            break
    if flag == 0 and ans < len_group:
        ans = len_group

print(ans)