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)

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,sse3,sse4,popcnt,abm,mmx")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define pb push_back
#define F first
#define S second
#define ins insert
#define mp make_pair
#define fo(i, n1, n, x) for(int i = n1; i <= n; i += x)
#define foo(i, n, n1, x) for(int i = n; i >= n1; i -= x)
#define bit __builtin_popcount
#define md (l + ((r - l) / 2))
#define all(x) x.begin(),x.end()
#define eb emplace_back
#define ub upper_bound
#define lb lower_bound
#define ios ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define file(s) if (fopen(s".in", "r")) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout)

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
typedef long double ld;

const int N = 2e5 + 11, mod = 1e9 + 7, mod2 = 998244353;
const int MAX = 3e5 + 11;
const int INF1 = 2e9 + 11;
const ll INF2 = 2e18 + 11;
const double INF3 = 1e8 + 11;
const int base = 500;
const int P = 31;
const int dx[] = {1, -1, 0, 0, 1, 1, -1, -1};
const int dy[] = {0, 0, 1, -1, 1, -1, 1, -1};
const double EPS = 1e-6;
const double PI = acos(-1.0);


template<typename T> using ordered_set = tree <T, null_type, less <T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T1, typename T2> inline void chmin(T1 &a, T2 b) { if (a > b) a = b; }
template<typename T1, typename T2> inline void chmax(T1 &a, T2 b) { if (a < b) a = b; }


typedef pair <int, int> pii;
typedef pair <ll, ll> pll;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int a,b,c;
main() {
    file("cowdate");
    ios;
    cin>>a>>b>>c;
    cout<<max(c-(a-b),0);
    return 0;
}
/**

*/