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)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define sz(x) ((int)x.size())
#define all(x) (x).begin(), (x).end()
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define mset(x,y) memset(x,y,sizeof(x))
#define rep(i,n) for(int i=0;i<int(n);i++)
#define per(i,n) for(int i=int(n)-1;i>=0;i--)
#define rng(i,a,b) for(int i=int(a);i<=int(b);i++)
#define gnr(i,b,a) for(int i=int(b);i>=int(a);i--)
using namespace std;
using namespace __gnu_pbds;
using ordered_set = tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>;
#ifdef LOCAL
#define dmp(...) _dmp(#__VA_ARGS__, __VA_ARGS__)
#else
#define dmp(...) (__VA_ARGS__)
#endif
template<class T> using vt=vector<T>;
template<class T> using vvt=vt<vt<T>>;
template<class TA,class TB> void chmax(TA&a,TB b){if(a<b)a=b;}
template<class TA,class TB> void chmin(TA&a,TB b){if(b<a)a=b;}
template<class TA,class TB>
ostream& operator<<(ostream& os,const pair<TA,TB>& p){
	return os<<"{"<<p.fi<<","<<p.se<<"}";
}
template<class T> ostream& operator<<(ostream& os,const vt<T>& v){
	os<<"{";for(auto& e:v)os<<e<<",";return os<<"}";
}
template<class TH> void _dmp(const char *sdbg, TH h){cout<<sdbg<<"="<<h<<endl;}
template<class TH, class... TA> void _dmp(const char *sdbg, TH h, TA...a){
while(*sdbg!=',')cout<<*sdbg++;cout<<"="<<h<<","; _dmp(sdbg+1, a...);
}
template<class T> void make_unique(vt<T>& v) {
	sort(all(v));v.erase(unique(all(v)),v.end());
}
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
int ri(){int x;cin>>x;return x;}
ll rll(){ll x;cin>>x;return x;}
vi rvi(int n){vi v;rep(i,n)v.pb(ri());return v;}
ll n,a,b;
void MAIN() {
	cin>>n>>a>>b;
	ll ans=9e18;
	if((b-a)%2) {
		chmin(ans,b-1);
		chmin(ans,n-a);
	} else {
		chmin(ans,(b-a)/2);
	}
	cout<<ans;
}
int main() {
	ios::sync_with_stdio(false);cin.tie(0);
	int T=1;
	//cin>>T;
	for(int tt=1;tt<=T;tt++) {
		MAIN();
	}
	return 0;
}