Problem:
Score : 400 points

Problem Statement

There are two persons, numbered 0 and 1, and a variable x whose initial value is 0.
The two persons now play a game.
The game is played in N rounds. The following should be done in the i-th round (1 \leq i \leq N):

Person S_i does one of the following:

Replace x with x \oplus A_i, where \oplus represents bitwise XOR.

Do nothing.

Person 0 aims to have x=0 at the end of the game, while Person 1 aims to have x \neq 0 at the end of the game.

Determine whether x becomes 0 at the end of the game when the two persons play optimally.

Solve T test cases for each input file.

Constraints

1 \leq T \leq 100

1 \leq N \leq 200

1 \leq A_i \leq 10^{18}

S is a string of length N consisting of 0 and 1.

All numbers in input are integers.

Input

Input is given from Standard Input in the following format.
The first line is as follows:

T

Then, T test cases follow.
Each test case is given in the following format:

N
A_1 A_2 \cdots A_N
S

Output

For each test case, print a line containing 0 if x becomes 0 at the end of the game, and 1 otherwise.

Sample Input 1

3
2
1 2
10
2
1 1
10
6
2 3 4 5 6 7
111000

Sample Output 1

1
0
0

In the first test case, if Person 1 replaces x with 0 \oplus 1=1, we surely have x \neq 0 at the end of the game, regardless of the choice of Person 0.

In the second test case, regardless of the choice of Person 1, Person 0 can make x=0 with a suitable choice.

Solution in Python (3.8.2) (judge verdict: Time Limit Exceeded)

def resolve():
    T = int(input())

    for t in range(T):
        n = int(input())
        A = list(map(int, input().split()))
        s = input()
        if s[-1] == "1":
            print(1)
            break
        p1 = [0]
        p2 = [0]
        c1 = [A[i] for i in range(n) if s[i] == "0"]
        c2 = [A[i] for i in range(n) if s[i] == "1"]

        def dfs1(x, n):
            if n == len(c1):
                p1.append(x)
                return
            dfs1(x, n + 1)
            dfs1(x ^ c1[n], n + 1)

        def dfs2(x, n):
            if n == len(c2):
                p2.append(x)
                return
            dfs2(x, n + 1)
            dfs2(x ^ c2[n], n + 1)

        dfs1(0, 0)
        dfs2(0, 0)
        # print(set(p1), set(p2))
        if len(set(p1) ^ set(p2)) == 0:
            print(0)
        else:
            print(1)

resolve()