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;
const int INF = 1e10;
const int MOD = 1e9 + 7;
#define dump(x) cout << #x << " = " << (x) << endl;
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define FORR(i, a, b) for (int i = (a); i <= (b); ++i)
#define pb push_back
#define all(v) (v).begin(), (v).end()
#define P pair<int, int>
#define V vector<int>
#define M map<int, int>
#define S set<int>
#define itn int
#define tin int
#define tni int
bool dbg = false;
int a, b, c;
void solve()
{
}

signed main()
{
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cin >> a >> b >> c;
    cout << max((c - (a - b)), 0) << endl;
    solve();
    return 0;
}