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:
from itertools import groupby, chain
from operator import itemgetter


def sweep(l):
    result = []
    s = set(l)
    while s:
        a = max(s)
        b = 1 << (a.bit_length() - 1)
        s = {t ^ a if t >= b else t for t in s}
        s.discard(0)
        result.append([a, b])

    return result


t = int(input())
buf = []
for _ in range(t):
    n = int(input())
    aaa = list(map(int, input().split()))
    s = input()

    if s[-1] == '1':
        buf.append(1)
        continue

    grp = [list(map(itemgetter(1), itr)) for c, itr in groupby(zip(s, aaa), key=itemgetter(0))]
    if s[0] == '0':
        grp = grp[1:]
    grp1 = grp[0::2]
    grp0 = grp[1::2]
    erasable = [False] * sum(len(g) for g in grp1)
    for i in range(len(grp1)):
        l1 = list(chain(*grp1[:i + 1]))
        l0 = list(chain(*grp0[i:]))
        s0 = sweep(l0)

        for a, b in s0:
            for i in range(len(l1)):
                if l1[i] & b:
                    l1[i] ^= a

        for j, c in enumerate(l1):
            if c == 0:
                erasable[j] = True

    if all(erasable):
        buf.append(0)
    else:
        buf.append(1)


print('\n'.join(map(str, buf)))