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: Runtime Error)

N = int(input())
if N == 1:
    print(1)
    exit()
A = [0 for _ in range(N+1)]
l = [0,1]
H = []
for i in range(2**N):
    S = str(bin(i)).zfill(N)
    H.append(S[2:].zfill(N))
IXY = []
for i in range(1,N+1):
    A[i] = int(input())
    if A[i] == 0:
        continue
    for j in range(A[i]):
        x,y = map(int,input().split())
        IXY.append([i,x,y])
ANS = []
for h in H:
    for ixy in IXY:
        flg = True
        i,x,y = ixy[0]-1,ixy[1]-1,ixy[2]
        if h[i] == "1":
            if h[x] != str(y):
                flg = False
                break
    if flg:l.append(ANS.append(h))
ans = 0
for a in ANS:
    ans = max(a.count("1"),ans)
print(ans)