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: Wrong Answer)

#include <bits/stdc++.h>
#include <type_traits>
using namespace std;

using ll = long long;

#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rrep(i, n) for (int i = 1; i <= (n); ++i)
#define drep(i, n) for (int i = (n)-1; i >= 0; --i)
#define ddrep(i, n) for (int i = n; i > 0; --i)
#define srep(i, s, t) for (int i = s; i < t; ++i)
#define ssrep(i, s, t) for (int i = s; i <= t; ++i)

#define rng(a) a.begin(), a.end()
#define pb push_back
#define eb emplace_back
#define fi first
#define se second

#define chmax(x, y) (x = max(x, y))
#define chmin(x, y) (x = min(x, y))

using pi = pair<int, int>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;
using ld = long double;

template <class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  os << "(" << p.first << "," << p.second << ")";
  return os;
}

template <class T> ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  rep(i, (int)v.size()) {
    if (i)
      os << ",";
    os << v[i];
  }
  os << "}";
  return os;
}

template <typename T, size_t S> void printArray(const T (&array)[S]) {
  for (auto val : array)
    std::cout << val << ", ";
  std::cout << "\n";
}

const int mod = 1e9 + 7;
const int inf = 1e9 + 5;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << std::setprecision(10);

  string s; std::cin >> s;
  int n = s.size();
  vi ans(n);
  rep(i, n) {
    int ni = i;
    while (ni+1 < n && s[i] == s[ni+1]) ni++;
    int cnt = ni - i + 1;
    if (s[i] == 'L') {
      ans[i] += (cnt + 1) / 2;
      if (i-1 >= 0) ans[i-1] += cnt / 2;
      else ans[i] += cnt / 2;
    } else {
      ans[ni] += (cnt + 1) / 2;
      if (ni+1 < n) ans[ni+1] += cnt / 2;
      else ans[ni] += cnt / 2;
    }
    i = ni;
  }
  rep(i, n) std::cout << ans[i] << (i+1 < n ? " " : "\n");
}