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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,n) for(ll i=0;i<n;++i)
#define Graph vector<vector<ll>>
constexpr ll INF=(1ll<<60);
constexpr ll mod=1000000007;
constexpr double pi=3.14159265358979323846;
template<typename T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;} return 0;}
template<typename T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;} return 0;}

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