# The XOR is simple enough. The trick is that they haven't given us the
# numbers to XOR, but their position in the list. To find the actual
# numbers, determine the number of numbers up to N bits long in the list.
# Count 0, since we can add any new bit to zero.
# Then there are 2 numbers with up to 1 bit, 3 with up to 2 bits, etc.
# To add the numbers with N bits, add the number of numbers with N-2 bits
# or less to the previous total (the numbers for N-1 bits).
# The result is the fibonacci sequence, since I just keep adding the last
# two numbers. And so the number I want is the Zeckendorf representation
# of the given index, reinterpreted as binary (repeatedly subtracting the
# largest Fibonacci number available from the index)
#
# I've got one case here that is failing, probably just barely failing the time limit.
# Let's try precalculating the powers as the small speedup needed.

def zeck(n):
    res = 0
    for i in range(f - 1, -1, -1):
        if n >= fib[i]:
            n -= fib[i]
            res += p2[i]
    return res

n = int(input())
nums = list(map(int, input().strip().split()))
fib = [1, 2, 3]
while fib[-1] < 10 ** 18:
    fib.append(fib[-1] + fib[-2])
f = len(fib)
p2 = []
for j in range(f):
    p2.append(2 ** j)
modulus = 10 ** 9 + 7
sol = 0
for j in range(n):
    sol = sol ^ zeck(nums[j])
print(sol % modulus)