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)

/*
Author : N_o_o_B
Created : December 28 2019 17:33:32 
*/
#include <bits/stdc++.h>
using namespace std;
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
*/
#define TRACE
#ifdef TRACE
#define trace(...) { cerr<<"[ "; __trace__(#__VA_ARGS__, __VA_ARGS__);}
#undef endl
template <typename Arg1,typename Arg2>
ostream& operator << (ostream& out, const pair<Arg1,Arg2> &x) {
    return out<<"("<<x.first<<","<<x.second<<")";
}
 
template <typename Arg1>
ostream& operator << (ostream& out, const vector<Arg1> &a) {
    out<<"[";for(const auto &x:a)out<<x<<",";return out<<"]";
}
 
template <typename Arg1>
ostream& operator << (ostream& out, const set<Arg1> &a) {
    out<<"[";for(const auto &x:a)out<<x<<",";return out<<"]";
}
 
template <typename Arg1,typename Arg2>
ostream& operator << (ostream& out, const map<Arg1,Arg2> &a) {
    out<<"[";for(const auto &x:a)out<<x<<",";return out<<"]";
}
 
template <typename Arg1>
void __trace__(const string name, Arg1&& arg1){
    cerr << name << " : " << arg1 << " ] " << endl;
}
 
template <typename Arg1, typename... Args>
void __trace__(const string names, Arg1&& arg1, Args&&... args){
    const string name = names.substr(0,names.find(','));
    cerr<<name<<" : "<<arg1<<" | ";
    __trace__(names.substr(1+(int)name.size()), args...);
}
#else
    #define trace(args...)
#endif
 
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<pair<int,int>> vii;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<pair<ll,ll>> vll;
typedef vector<vl> vvl;
//typedef tree<pii, null_type, less<pii>, rb_tree_tag, tree_order_statistics_node_update> oset;
#define fori(i,n) for(int i=0;i<n;i++)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define ford(i,n)  for(int i=n-1;i>=0;i--)
#define trav(x,a) for(auto& x:a)
#define all(x) x.begin(),x.end()
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define fi first
#define se second
clock_t time_p=clock();
void time_taken()
{
    time_p=clock()-time_p;
    cerr<<"Time Taken : "<<(float)(time_p)/CLOCKS_PER_SEC<<"\n";
}
//const ll mod=998244353;
const ll mod=1e9+7;
const ll INF=1e18;

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);cout.tie(NULL);
    cout.precision(12);cout << fixed;

    ll n,a,b;
    cin >> n >> a >> b;
    if(a > b) swap(a,b);
    if((b-a)%2 == 0){
        cout << (b-a)/2 << endl;
    }
    else{
        cout << min(n-a,b-1) << endl;
    }



    time_taken();
    return 0;
}