## Leap Year

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.

- The year can be evenly divided by 100, it is NOT a leap year, unless:

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<=10^5

**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.

**Code:**

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

**Disclaimer: **These problems are originally created and published by HackerRank, we only provide solutions to those problems.Hence, doesn’t guarantee the truthfulness of the problem. This is only for information purposes.