/*
-For subtask 1 just repeatedly find the next biggest and smallest number.
-For subtask 2, find the biggest and the smallest number (cost will be N+1);
-Now notice that the solution will be at least (r-l+1-n)/(n-1) rounded up.
-We can split the search space into parts of that size and query for the biggest and smallest numbers.
-We don't care if we have some gap/numbers inbetween the biggest and smallest numbers of a part since the gap can not be big enough to be the biggest.
*/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
#include "gap.h"
#define ll long long
#define pb push_back
#define sz(x) (int)(x).size()
#define mp make_pair
#define f first
#define s second
#define all(x) x.begin(), x.end()
#define D(x) cerr << #x << " is " << (x) << "\n";
using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>; ///find_by_order(),order_of_key()
template<class T1, class T2> ostream& operator<<(ostream& os, const pair<T1,T2>& a) { os << '{' << a.f << ", " << a.s << '}'; return os; }
template<class T> ostream& operator<<(ostream& os, const vector<T>& a){os << '{';for(int i=0;i<sz(a);i++){if(i>0&&i<sz(a))os << ", ";os << a[i];}os<<'}';return os;}
template<class T> ostream& operator<<(ostream& os, const set<T>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T> ostream& operator<<(ostream& os, const multiset<T>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T1,class T2> ostream& operator<<(ostream& os, const map<T1,T2>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
ll findGap(int t,int n)
{
vector<ll> arr;
ll l=0,r=1e18,a,b;
if(t==1)
{
for(int i=0;i<(n+1)/2;i++)
{
MinMax(l,r,&a,&b);
arr.pb(a);
arr.pb(b);
l=a+1;
r=b-1;
}
}
else
{
MinMax(l,r,&l,&r);
ll deo=(r-l+1-n)/(n-1);
if((r-l+1-n)%(n-1))
deo++;
arr.pb(l);
arr.pb(r);
l++;
r--;
for(int i=0;i<n-1;i++)
{
ll le=l+(deo+1000)*i,ri=min(r,l+(deo+1000)*(i+1)-1);
if(le>ri)
continue;
MinMax(le,ri,&a,&b);
if(a!=-1)
arr.pb(a),arr.pb(b);
}
}
sort(all(arr));
ll sol=0;
for(int i=0;i<(int)arr.size()-1;i++)
sol=max(sol,arr[i+1]-arr[i]);
return sol;
}
