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:


def I(): return int(input())
def MI(): return map(int, input().split())
def LI(): return list(map(int, input().split()))

def main():
    mod=10**9+7
    T=I()
    M=60
    


    
    def pipo(x):
        cnt=0
        for i in range(M):
            if x==0:
                return cnt
            else:
                x=x//2
                cnt+=1
    

        
    
    for _ in range(T):
        N=I()
        A=LI()
        S=list(input())
        #後ろから見ていき，S=0の数だけで作れない数が(S=1の時に)出てきたらNG
        Vec=[]
        flag=1
        for i in range(N-1,-1,-1):
            #print(i,S[i]) 
            if S[i]=="0":
                if Vec==[]:
                    Vec.append((A[i],pipo(A[i])))
                else:
                    now=A[i]
                    p=pipo(A[i])
                    for v in Vec:
                        if v[1]==p:
                            now=now^v[0]
                            p=pipo(now)
                    if now!=0:
                        Vec.append((now,pipo(now)))
                        Vec.sort(reverse=True)
                            
                
            else:
                #print(Vec)
                now=A[i]
                for v in Vec:
                    if v[1]==pipo(now):
                        now=now^v[0]
                if now!=0:
                    flag=0

                    
                
        if flag:
            print(0)
        else:
            print(1)
            
            
        
        
        
        

main()