Write a function that receives two strings as parameter. This strings are in the following format of date: `YYYY/MM/DD`.
Your job is: Take the `years` and calculate the difference between them.

Examples:
```
'1997/10/10' and '2015/10/10' -> 2015 - 1997 = returns 18 
'2015/10/10' and '1997/10/10' -> 2015 - 1997 = returns 18
```
At this level, you don't need validate months and days to calculate the difference.

Difficulty: introductory

Starter code:

def how_many_years (date1,date2):
	

Solutions:
def how_many_years (date1,date2):
  return abs(int(date1.split('/')[0]) - int(date2.split('/')[0]))


def how_many_years (date1,date2):
    return abs(int(date1[:4]) - int(date2[:4]))

def how_many_years (date1,date2):
    year = lambda d: int(d[:4])
    return abs(year(date1) - year(date2))

def how_many_years (date1, date2):
    get_year = lambda date: int(date.split('/')[0])
    return abs(get_year(date1) - get_year(date2))

how_many_years = lambda d1,d2: abs(int(''.join(d1[:4])) - int(''.join(d2[:4])))

def how_many_years (*dates):
    years = [int(date.split('/')[0]) for date in dates]
    return abs(years[0]-years[1])

from datetime import datetime

def how_many_years (*dates):
    d1,d2 = [datetime.strptime(s, "%Y/%m/%d") for s in dates]
    return abs(d2.year - d1.year)
    


HowManyYears = how_many_years = lambda *a: abs(int.__sub__(*(int(d.split('/')[0]) for d in a))) 

def how_many_years (date1,date2):
    a,b = map(int,map(lambda x: x.split('/')[0],(date1,date2)))
    return abs(b-a)

def how_many_years (d1,d2):
    a=d1[0:4]
    b=d2[0:4]
    print( (int(a) , int(b)))
    return abs(int(a) - int(b))