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)

// lcmとか__builtin_popcountとかはg++ -std=c++17 default.cppみたいなかんじで

#include <bits/stdc++.h>
#define mod 1000000007
#define INF LLONG_MAX
#define ll long long
#define ln cout<<endl
#define Yes cout<<"Yes"<<endl
#define NO cout<<"NO"<<endl
#define YES cout<<"YES"<<endl
#define No cout<<"No"<<endl
#define REP(i,m,n) for(ll i=(ll)(m);i<(ll)(n);i++)
#define rep(i,n) REP(i,0,n)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
using namespace std;
ll dx[4]={1,0,-1,0};
ll dy[4]={0,1,0,-1};

int main() {
    cin.tie(0);
   	ios::sync_with_stdio(false);

    ll  a,b,c,d,e,n,k,maxi=0,f=0,mini=INF,sum=0,q;
	string str,stra,straa;
    ll x,y,z;
    char ca,cb,cc;
    cin>>a>>b>>c;
    if((a-b)-c<0) cout<<c-(a-b)<<endl;
    else cout<<0<<endl;

    return 0;
}