# Write a function

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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

**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.

**Input Format**

Read year, the year to test.

**Constraints**

1900 ≤ year ≤ 10⁵

**Output Format**

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

**Sample Input 0**

`1990`

**Sample Output 0**

`False`

**Explanation 0**

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

Solution(Python3):

def is_leap(year):leap = False

if year %4 == 0 and (year % 400 == 0 or year % 100 == 0):

leap = True

return leapyear = int(input())print(is_leap(year))