Help please! Due in 25 minutes!

1 # Write a funetion named "count_ evenly divisible' that takes 4 arguments 2 # Each argument will be an int value 3 The first two arguments will be a start value and an end value. 4 # The start value will always be less than the end value. the arguments will be two divisors. 6 the function should determine and return the number of integer values between the start and end values 7 f that are evenly divisible by either the first or the second divisor value. 9 # Examples : 10 count-evenly-divisible (1, io, 2, 3) should return 7 11 # 12 # evenly divisible by or evenly divisible by 3. 13 14 # 15 # count-evenly-divisible(10, 30, 5, 7) should return 8, 16 because there 2, 3,-4, 6, 8, 9, 10 are all-either evenly.divisible by 2 or evenly divisible by 3 because there are 8 integer values between 10 and 30 that are either 10, 14, 15, 20, 21, 25, 28, and 30 are all ither evenly divisible by 5 or evenly divisible by7 because there are 5 integer values between 5 and 23 that are either 6, 11, 12, 18, and 22 are al either evenly divisible or evenly divisible by 11 are 7 integer values between 1 and 10 that are either 17 evenly diviaible by 5 or evenly diviaible by 7 18 19 20 count-evenly-divisible(5, 23, 6, 11) should returns, 21 # 22 # evenly divisible by 6 or evenly divisible by 11. 23 24 25 # Remember that you can use the modulus (remainder) operator (5) to get the remainder of a division operation, 26 for example: 27 # 20 7 evaluates to 6 because 7 goes into 20 to times with 6 remainin 29-21 % 7 evaluates to 0 because 7 goes into 21 three times with 0 remaining 30 # 31 21 is evenly divisible by 7, but 20 and 22 are not evenly divisible by 7 32 # 33 # DO NOT include any code other than your function definit ion in your submission