Problem:
Score : 300 points

Problem Statement

2N players are running a competitive table tennis training on N tables numbered from 1 to N.

The training consists of rounds.
In each round, the players form N pairs, one pair per table.
In each pair, competitors play a match against each other.
As a result, one of them wins and the other one loses.

The winner of the match on table X plays on table X-1 in the next round,
except for the winner of the match on table 1 who stays at table 1.

Similarly, the loser of the match on table X plays on table X+1 in the next round,
except for the loser of the match on table N who stays at table N.

Two friends are playing their first round matches on distinct tables A and B.
Let's assume that the friends are strong enough to win or lose any match at will.
What is the smallest number of rounds after which the friends can get to play a match against each other?

Constraints

2 \leq N \leq 10^{18}

1 \leq A < B \leq N

All input values are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

Print the smallest number of rounds after which the friends can get to play a match against each other.

Sample Input 1

5 2 4

Sample Output 1

1

If the first friend loses their match and the second friend wins their match, they will both move to table 3 and play each other in the next round.

Sample Input 2

5 2 3

Sample Output 2

2

If both friends win two matches in a row, they will both move to table 1.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

// clang-format off
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define stoi stoll
using pii = pair<int,int>;
using vi = vector<int>;
using vvi = vector<vi>;
#define all(c) begin(c), end(c)
#define rall(c) rbegin(c), rend(c)
#define fore(x, c) for(auto &&x : c)
#define rep(i, a, n) for(int i = a, i##len = (int)(n); i < i##len; ++i)
#define rrep(i, a, n) for(int i = (int)(n - 1); i >= a; --i)
#define sz(c) ((int)c.size())
#define contains(c, x) (c.find(x) != end(c))
#define inseg(l, x, r) ((l) <= (x) && (x) < (r))
#define dump(...)
#define pb push_back
#define eb emplace_back
#define _ 0
const signed INF_ = 1001001001; const long long INF = 1001001001001001001LL;
const int DX[9] = {0,1,0,-1,1,1,-1,-1,0}, DY[9] = {-1,0,1,0,-1,1,1,-1,0};
template<class T> ostream& operator<<(ostream &os, const vector<T> &v) {
  for (auto i = begin(v); i != end(v); i++) os<<*i<<(i==end(v)-1?"":" ");return os;}
template<class T> istream& operator>>(istream &is, vector<T> &v) {
  for (auto i = begin(v); i != end(v); i++) is>>*i;return is;}
template<class T,class U> istream& operator>>(istream &is, pair<T,U> &p) {
  is>>p.first>>p.second;return is;}
template<class T, class U> bool chmax(T &a, const U &b){return a<b?a=b,1:0;}
template<class T, class U> bool chmin(T &a, const U &b){return a>b?a=b,1:0;}
template <class T> void psum(T& c) {partial_sum(begin(c), end(c), begin(c));}
template<class T> using heap = priority_queue<T, deque<T>, greater<T>>;
struct before_main_function {
  before_main_function() {
      cin.tie(nullptr); ios::sync_with_stdio(0);
      cout << setprecision(15) << fixed;
  }
} before_main_function;
// clang-format on
//------------------8<------------------------------------8<--------------------

signed main() {
  int N;
  cin >> N;
  int A, B;
  cin >> A >> B;
  if ((B - A) % 2 == 0) {
    cout << (B - A) / 2 << endl;
  } else {
    int ans = INF;
    chmin(ans, B - 1);
    chmin(ans, N - A);
    cout << ans << endl;
  }
  return (0 ^ _ ^ 0);
}