# Python solution to "Divisor Game" from 2025 CS@UCF Summer camp

# Note that for any number n > 2, n - 1 is not a divisor of n. Furthermore,
# n - (n - 1) is the only expression of the form (n - x) that evaluates to 1, the value
# that results in a win for the person who could force their opponent to play when n = 1
# Therefore, the only state that leads directly to a win is to play whenever n = 2
# Since all divisors of an odd number are odd, and an odd number subtracted from an odd
# number creates an even number, if we can start our turn with an even number, we can subtract 1 from it,
# which ensures that every time it is our turn we play on an even number,
# and since the value is always decreasing, the even number we play on will eventually be 2, at which point
# we can subtract 1 from the value to force our opponent into a loss.

test_count = int(input())

for _ in range(test_count):
    # To test if a number is odd, we can check if the least bit (with a value of 1) is turned on
    # If the 1 bit is turned on, the value is odd, otherwise it is even
    if((int(input()) & 1) == 0):
        print("Grace")
    else:
        print("Christian")