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 (3.8.2) (judge verdict: Wrong Answer)

def main():
    N = int(input())
    A_lst = [0] * N
    xy_lst = [[]] * N
    ans = 0

    for i in range(N):  # 入力
        A = int(input())
        A_lst[i] = A
        xy_lst[i] = [[]] * A

        for j in range(A):
            x, y = map(int, input().split())
            xy_lst[i][j] = [x, y]  # 3次元配列 i人目のj番目の証言の[x,y]

    for i in range(2 ** N):  # ビット全探索
        honest_unkind = [0] * N
        flag = True

        for j in range(N):  # 1つ状態を決める
            if (i >> j) & 1 == 1:
                honest_unkind[j] = 1  # 1:honest, 0:unkind

        for j in range(N):  # 1人ずつ証言を検証する
            for k in range(A_lst[j]):  # 各人の証言がk個
                if honest_unkind[j] == 1:  # j番目の人がhonestだったら
                    if xy_lst[j][k][1] == 0 and honest_unkind[xy_lst[j][k][0] - 1] == 1:
                        flag = False  # 矛盾したら終了
                        break
                    if xy_lst[j][k][1] == 1 and honest_unkind[xy_lst[j][k][0] - 1] == 0:
                        flag = False
                        break
                if honest_unkind[j] == 0:  # j番目の人がunkindだったら
                    if xy_lst[j][k][1] == 0 and honest_unkind[xy_lst[j][k][0] - 1] == 0:
                        flag = False
                        break
                    if xy_lst[j][k][1] == 1 and honest_unkind[xy_lst[j][k][0] - 1] == 1:
                        flag = False
                        break

        if flag == True:  # 全員の証言が矛盾しなかった場合
            cnt = sum(honest_unkind)
            if cnt > ans:
                ans = cnt

    print(ans)


if __name__ == "__main__":
    main()