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 <algorithm>
#include <bits/stdc++.h>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
using namespace std;

#define rep(X, S, E) for (int(X) = (S); (X) < (E); ++(X))
#define rrep(X, S, E) for (int(X) = (E)-1; (X) >= (S); --(X))
#define itrep(X, Y) for (auto(X) = (Y).begin(); (X) != (Y).end(); (X)++)
#define all(X) (X).begin(), (X).end()
#define sortDecending(X) sort(all(X), greater<ll>()) //降順
#define sortAscending(X) sort(all(X))                //昇順
#define pb push_back
#define mp make_pair
#define fi first
#define sc second
#define print(x) cout << x << endl
#define printDouble(x) cout << fixed << setprecision(13) << x << endl
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<bool> vb;
typedef priority_queue<ll, vl> decendingQueue;              //降順
typedef priority_queue<ll, vl, greater<ll>> ascendingQueue; //昇順
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const ll INF = 9 * 1e18;
const ll MOD = 1e9 + 7;

long long gcd(long long m, long long n) {
  if (m < n)
    return gcd(n, m);
  if (n == 0)
    return m;
  return gcd(n, m % n);
}

long long lcm(long long m, long long n) {
  // m * nでlong型のオーバフローを発生させないため、先に割り算から行う
  return m * (n / gcd(m, n));
}

// a + b + cをprimeで割った値を返す
long long addMod(long long a, long long b) { return (a + b) % MOD; }

long long minusMod(long long a, long long b) { return (a + MOD - b) % MOD; }

long long multipleMod(long long a, long long b) { return (a * b) % MOD; }

vector<long long> SieveOfEratosthenes(int max) {
  vector<long long> sieve;
  vector<long long> primes;

  for (int i = 1; i < max + 1; ++i) {
    sieve.push_back(i);
  }

  sieve[0] = 0;
  for (int i = 2; i < max + 1; ++i) {
    if (sieve[i - 1] != 0) {
      primes.push_back(sieve[i - 1]);
      for (int j = 2 * sieve[i - 1]; j < max + 1; j += sieve[i - 1]) {
        sieve[j - 1] = 0;
      }
    }
  }
  return primes;
}

class Combination {
private:
  vector<long long> fac_;
  vector<long long> finv_;
  vector<long long> inv_;
  long long prime_;

public:
  Combination(long long n, long long prime) {
    fac_ = vector<long long>(n + 1);
    finv_ = vector<long long>(n + 1);
    inv_ = vector<long long>(n + 1);
    prime_ = prime;

    fac_[0] = fac_[1] = 1;
    finv_[0] = finv_[1] = 1;
    inv_[1] = 1;
    for (long long i = 2; i <= n; i++) {
      fac_[i] = fac_[i - 1] * i % prime_;
      inv_[i] = prime_ - inv_[prime_ % i] * (prime_ / i) % prime_;
      finv_[i] = finv_[i - 1] * inv_[i] % prime_;
    }
  }

  // nCk
  long long getCombination(long long n, long long k) {
    if (n < k)
      return 0;
    if (n < 0 || k < 0)
      return 0;

    return fac_[n] * (finv_[k] * finv_[n - k] % prime_) % prime_;
  }
};

class UnionFindTree {
private:
  vector<int> par;
  vector<int> rnk;
  vector<int> siz;

public:
  UnionFindTree(int n) {
    par.assign(n, -1);
    rnk.assign(n, -1);
    siz.assign(n, -1);

    for (int i = 0; i < n; ++i) {
      par[i] = i;
      rnk[i] = 0;
      siz[i] = 1;
    }
  }

  int find(int x) {
    if (par[x] == x)
      return x;
    else
      return par[x] = find(par[x]);
  }

  bool same(int x, int y) { return find(x) == find(y); }

  void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y)
      return;
    if (rnk[x] < rnk[y]) {
      par[x] = y;
      siz[y] += siz[x];
    } else {
      par[y] = x;
      siz[x] += siz[y];
      if (rnk[x] == rnk[y])
        ++rnk[x];
    }
  }

  int size(int x) {
    x = find(x);
    return siz[x];
  }
};

class Edge {
public:
  ll from;
  ll to;
  ll cost;

  Edge() {}
  Edge(ll from, ll to, ll cost) {
    this->from = from;
    this->to = to;
    this->cost = cost;
  }

  bool operator<(const Edge &edge) const {
    return cost < edge.cost; //昇順
  }
};

class Graph {
public:
  ll nodes; // ノード数
  vector<Edge> edges;

  Graph() {}
  Graph(ll nodes) { this->nodes = nodes; }
  void addEdge(ll from, ll to, ll cost) {
    this->edges.push_back(Edge(from, to, cost));
  }
};

// クラスカル法
// 連結グラフの最小全域木を求める
class Kruskal {
private:
  Graph graph;
  vector<Edge> MinimumSpanningTree;
  ll minimumCost;
  void searchMinimumSpanningTree() {
    UnionFindTree uf(graph.nodes);
    sort(all(graph.edges));
    itrep(edge, graph.edges) {
      if (!uf.same(edge->from, edge->to)) {
        uf.unite(edge->from, edge->to);
        MinimumSpanningTree.push_back(*edge);
      }
    }
  }

public:
  Kruskal(Graph graph) { this->graph = graph; }
  ll getMinimumSpanningTreeCost() {
    searchMinimumSpanningTree();
    ll cost = 0;
    itrep(it, MinimumSpanningTree) { cost += it->cost; }
    return cost;
  }
};

// ダイクストラ法 O((E+V)logV)
// 最小経路問題を解くためのアルゴリズム。辺の重みに負数を含む場合は利用不可
class Dijkstra {
private:
  Graph graph;
  map<ll, vector<Edge>> fromPaths;
  vl distances;
  vl srcs;

public:
  Dijkstra(Graph graph) {
    this->graph = graph;
    itrep(edge, graph.edges) {
      fromPaths[edge->from].push_back(*edge);
      fromPaths[edge->to].push_back(Edge(edge->to, edge->from, edge->cost));
    }
  }
  void searchMinimumPathFrom(ll src) {
    // 複数回呼ばれる度に計算する
    this->distances = vl(graph.nodes + 1, INF);
    this->srcs = vl(graph.nodes + 1, INF);

    priority_queue<ll> pq;
    distances[src] = 0;
    srcs[src] = -1;

    pq.push(src);
    while (!pq.empty()) {
      int u = pq.top();
      pq.pop();

      itrep(edge, fromPaths[u]) {
        int v = edge->to;
        int w = edge->cost;
        if (distances[v] > distances[u] + w) {
          distances[v] = distances[u] + w;
          srcs[v] = u;
          pq.push(v);
        }
      }
    }
  };
  ll getDistance(ll n) { return this->distances[n]; }
  ll getFrom(ll n) { return this->srcs[n]; }
};

// ベルマンフォード O(|V||E|)
// 非負コストが含まれていても最短経路問題を解くためのアルゴリズム。閉路の検出も可能
class BellmanFord {
private:
  Graph graph;
  // 閉路が含まれるかは個々のノードごとに管理する必要あり
  vector<bool> hasNegativeCycles;
  vector<ll> distances;

public:
  BellmanFord(Graph graph) {
    this->graph = graph;
    this->distances = vector<ll>(this->graph.nodes + 1, INF);
    this->hasNegativeCycles = vector<bool>(this->graph.nodes + 1, false);
  }
  void searchMinimumPathFrom(ll src) {
    this->distances[src] = 0;

    for (ll i = 0; i < graph.nodes - 1; i++) {
      itrep(edge, graph.edges) {
        ll u = edge->from;
        ll v = edge->to;
        ll w = edge->cost;
        if (this->distances[u] + w < this->distances[v]) {
          this->distances[v] = this->distances[u] + w;
        }
      }
    }

    itrep(edge, graph.edges) {
      ll u = edge->from;
      ll v = edge->to;
      ll w = edge->cost;
      if (this->distances[u] + w < this->distances[v]) {
        this->hasNegativeCycles[v] = true;
      }
      if (this->hasNegativeCycles[u] == true) {
        this->hasNegativeCycles[v] = true;
      }
    }
  }
  ll getDistance(ll n) { return this->distances[n]; }
  bool hasNegativeCycle(ll n) { return this->hasNegativeCycles[n]; }
};

void solve(long long A, long long B, long long C) {
  ll ans = max(0ll, B + C - A);
  print(ans);
}

int main() {
  long long A;
  scanf("%lld", &A);
  long long B;
  scanf("%lld", &B);
  long long C;
  scanf("%lld", &C);
  solve(A, B, C);
  return 0;
}