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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define REP(i,n) for(int64_t i=0;i<(n);i++)
#define ALL(a)  (a).begin(),(a).end()
#define SORT(c) sort((c).begin(),(c).end())
#define REV(c) reverse((c).begin(),(c).end())
#define YES cout << "YES" << endl;
#define NO cout << "NO" << endl;
#define Yes cout << "Yes" << endl;
#define No cout << "No" << endl;
#define VIN(n,v) for(int64_t i=0;i<(n);i++){cin >> (v).at(i);}

using namespace std;
typedef int64_t I;
typedef pair<I,I> P;
typedef vector<I> V;

I MOD = pow(10,9)+7;

// 最大公約数
I gcd(I a, I b) {
  return b != 0 ? gcd(b, a % b) : a;
}

// 最小公倍数
I lcm(I a, I b) {
  return a * b / gcd(a, b);
}

int ctoi(char c){
  return c - '0';
}

int main(){
  I a,b,c;
  cin >> a >> b >> c;
  int x = c - (a -b);
  int t = max(x, 0);
  cout << t << endl;
  return 0;
}