# Arup Guha
# 5/31/2012
# Determine if a year is a leap year or not.
# Edited on 6/1/2026 to print out proper tense.

def main():

    # The current year.
    CURYEAR = 2026

    # Get the year from the user.
    year = int(input("Please enter a year.\n"))

    # Assume we have a leap year, by default.
    result = True

    # If it's not divisible by 4, it's not a leap year.
    if year%4 != 0:
        result = False

    # Also, screen for 1700, 1800, 1900, etc. which are also not leap years.
    elif year%100 == 0 and year%400 != 0:
        result = False

    # This always prints.
    print(year, end=" ")
    
    # Set up if based on year.
    if year < CURYEAR:

        # In the past.
        print("was", end=" ")

        # Insert word in this case.
        if not result:
            print("not", end=" ")
        print("a leap year.")
        
    elif year == CURYEAR:
        
        # Current.
        print("is", end=" ")

        # Insert word in this case.
        if not result:
            print("not", end=" ")
        print("a leap year.")
        
    else:

        # In the future.
        print("will", end=" ")

        # Insert word in this case.
        if not result:
            print("not", end=" ")
        print("be a leap year.")
    


main()