Problem

An extra day is added to the calendar almost every four years as February 29, and the day is called a leap day. It corrects the calendar for the fact that our planet takes approximately 365.25 days to orbit the sun. A leap year contains a leap day.

In the Gregorian calendar, three conditions are used to identify leap years:

• The year can be evenly divided by 4, is a leap year, unless:
• The year can be evenly divided by 100, it is NOT a leap year, unless:
• The year is also evenly divisible by 400. Then it is a leap year.

This means that in the Gregorian calendar, the years 2000 and 2400 are leap years, while 1800, 1900, 2100, 2200, 2300 and 2500 are NOT leap years. Source

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Task

Given a year, determine whether it is a leap year. If it is a leap year, return the Boolean True, otherwise return False.

Note that the code stub provided reads from STDIN and passes arguments to the is_leap function. It is only necessary to complete the is_leap function.

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Input Format

Read , the year to test.

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Output Format

The function must return a Boolean value (True/False). Output is handled by the provided code stub.

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Sample Input 0

1990

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Sample Output 0

False

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Explanation 0

1990 is not a multiple of 4 hence it's not a leap year.

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Solution

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def is_leap(year):
    leap = False
    
    if year%4==0 and year%100 !=0:
        leap =True
    elif year%400==0:
        leap =True
    
    return leap