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)

#define _USE_NATH_DEFINES
#include<cstdio>
#include <iostream>
#include<sstream>
#include<fstream>
#include<iomanip>
#include<algorithm>
#include<cmath>
#include<complex>
#include <string>
#include<vector>
#include<list>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<bitset>
#include<numeric>
#include<limits>
#include<climits>
#include<cfloat>
#include<functional>
#include<iterator>
#include <random>

using namespace std;
#define REP(i,m,n) for(int i=(int)m ; i < (int) n ; ++i )
#define rep(i,n) REP(i,0,n)
using ll = long long;
#define EPS 1e-9

int main() {
	int a, b, c;
	cin >> a >> b >> c;
	if (c - (a - b) < 0) cout << 0 << endl;
	else cout << c - (a - b) << endl;
	return 0;
}