def counterGame(n):
    turns = 0
    while n > 1:
        # Check if n is a power of 2
        if (n & (n - 1)) == 0:
            # If it's a power of 2, divide by 2
            n >>= 1  # Equivalent to n = n // 2
        else:
            # If not a power of 2, find the largest power of 2 less than n
            # This can be found by taking 2 raised to the power of (n.bit_length() - 1)
            # n.bit_length() gives the number of bits required to represent n,
            # which is equivalent to floor(log2(n)) + 1.
            # So, the largest power of 2 less than n is 2^(floor(log2(n))).
            # We can get this using a bit shift: 1 << (n.bit_length() - 1)
            largest_power_of_2 = 1 << (n.bit_length() - 1)
            n -= largest_power_of_2
        turns += 1

    # If the number of turns is odd, Louise wins (since she starts)
    # If the number of turns is even, Richard wins
    if turns % 2 == 1:
        return "Louise"
    else:
        return "Richard"