Problem:
Score : 100 points

Problem Statement

We have two bottles for holding water.

Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.

Bottle 2 contains C milliliters of water.

We will transfer water from Bottle 2 to Bottle 1 as much as possible.

How much amount of water will remain in Bottle 2?

Constraints

All values in input are integers.

1 \leq B \leq A \leq 20

1 \leq C \leq 20

Input

Input is given from Standard Input in the following format:

A B C

Output

Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.

Sample Input 1

6 4 3

Sample Output 1

1

We will transfer two milliliters of water from Bottle 2 to Bottle 1, and one milliliter of water will remain in Bottle 2.

Sample Input 2

8 3 9

Sample Output 2

4

Sample Input 3

12 3 7

Sample Output 3

0

Solution in Ruby (2.3.3) (judge verdict: Accepted)

def get_i() #空白区切の入力を数値(整数)の配列で返す
  return gets.chomp.split(" ").map(&:to_i)
end
def get_f() #空白区切の入力を数値(実数)の配列で返す
  return gets.chomp.split(" ").map(&:to_f)
end
def get() #空白区切の入力を文字列の配列で返す
  return gets.chomp.split(" ")
end
def get_nsp() #入力されたものを一文字ずつに区切った文字列の配列で返す
  return gets.chomp.split("")
end
def yn_judge(bool,y="Yes",n="No") #boolに真偽を投げることで、trueならy、falseならnの値を出力する
  return bool ? y : n 
end
def array(size1,init=nil,size2=-1) #size2に二次元配列時の最初の要素数、size1に次の配列の大きさ、initに初期値を投げることでその配列を返す
  if size2==-1
    return Array.new(size1){init}
  else
    return Array.new(size2){Array.new(size1){init}}
  end
end

A,B,C=get_i
puts C-(A-B)<0 ? 0 : C-(A-B)