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>
// #include <iostream>
// #include <vector>
// #include <map>
// #include <algorithm>
using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())

template<typename T>
T mod(T a, T b){
    return (a + b) % b;
}

template<typename T>
T gcd(T a, T b){
    if (b == 0) return a;
    return gcd(b, a % b);
}

ll lcm(ll a, ll b){
    return a * b / gcd(a,b);
}

// n を素因数分解
map<ll, ll> factor(ll n){
    map<ll, ll> mp;
    for (ll i = 2; i*i <= n; i++) {
        while (n % i == 0) {
            mp[i]++;
            n /= i;
        }
    }

    if (n > 1) {
        mp[n]++;
    }

    return mp;
}

// 文字列に "hoge" が含まれているか
// string str = "hogepiyo";
// if (str.find("hoge") != string::npos) {
//     cout << "文字列が含まれています" << endl;
// }

// char to string
// char c = 'a';
// string str = {c};

void solve(){
    int A, B, C;
    cin >> A >> B >> C;

    cout << max(0, C - (A-B)) << endl;

    return;
}

int main(int argc, char const* argv[])
{
    solve();
    return 0;
}