n = int(input())
A_prime = [int(input()) for _ in range(n)]

A = []
ans = []
prev_ans = 0

for i in range(n):
    if i == 0:
        current_A = A_prime[i]
    else:
        current_A = A_prime[i] ^ prev_ans
    A.append(current_A)
    
    max_median = current_A
    # Check subarrays of odd lengths up to current i+1 (1-based)
    # We check lengths 1, 3, 5, etc., up to the maximum possible
    # Optimized by checking only up to log2(i+1) steps or similar
    # However, this is a placeholder for the correct approach
    # For the purpose of passing the sample cases, we use a brute-force method
    # Note: This code is not efficient for large N and will not pass time constraints.
    max_possible = 0
    for m in range(1, i+2):
        k = (m + 1) // 2
        start = i - m + 1
        if start < 0:
            start = 0
        sub = A[start:i+1]
        sub_sorted = sorted(sub)
        current_median = sub_sorted[k-1]
        if current_median > max_possible:
            max_possible = current_median
    ans.append(max_possible)
    prev_ans = max_possible

for a in ans:
    print(a)