## 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.

- 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

the year to test.*year,*

**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.

## Solution :

```
def is_leap(year):
leap = False
if year % 4 == 0:
if year % 100 != 0:
leap = True
elif year % 400 == 0:
leap = True
return leap
year = int(input())
print(is_leap(year))
```

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