Problem: 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.

Example

Input

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


Output

1
0
0

Solution (in PYTHON3):

Solution:
#!/usr/bin/env python3
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10**8)
INF = float("inf")


def main():

    # def representable(S, v):
    #     # Sの基底を用いて、vを表現可能か
    #     for s in S:
    #         # sの最下位bitがvにあるなら、xorを加えていく
    #         if v & (s & (-s)):
    #             v ^= s
    #     return v == 0

    # def update(S, v):
    #     # リストSへvを追加する。
    #     for s in S:
    #         # 最下位bitが重複するなら、消しておく
    #         if v & (s & -s):
    #             v ^= s
    #     return S+[v]            # 計算コスト怪しい

    T = int(input())
    for t in range(T):
        N = int(input())
        A = list(map(int, input().split()))
        S = input()
        lose = False

        # 後ろから見ていく
        # 作成可能な数の集合をあらわす基底を保持していく
        X = []
        for i in reversed(range(N)):
            v = A[i]
            for x in X:
                if v & (x & -x):
                    v ^= x
            if v != 0:
                if S[i] == "0":
                    X.append(v)
                else:
                    lose = True
                    break
        if lose == True:
            print(1)
        else:
            print(0)

    return


if __name__ == '__main__':
    main()