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>
using namespace std; 
#define ll long long int
#define pb push_back
#define mp make_pair
#define pll pair<ll,ll>
#define INF LLONG_MAX
#define PI acos(-1)
const ll MOD = 1e9 + 7;
#define fir first
#define sec second
#define all(x) x.begin(),x.end()
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(c,a,b) for(ll c=a;c<b;c++)
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define min_heap priority_queue<pll,vector<pll>,greater<pll> >
ll dx4[]={-1,1,0,0};
ll dy4[]={0,0,-1,1};
ll dx8[]={-1,-1,-1,0,1,1,1,0};
ll dy8[]={-1,0,1,1,1,0,-1,-1};
ll a[200][200];
int main()
{
  int a,b,c;
  cin>>a>>b>>c;
  cout<<max(c-a+b,0);
}