For this Question: You are given a permutation P of size N. You are also given an integer M. You have to cut the given permutation into M subsegments so that each element belongs to exactly one subsegment. After that you can reorder the subsegments in any order, however, elements belonging to the same subsegment should stay together, and in the same order.

What is the lexicographically smallest permutation you can achieve after reordering?The first line contains two space-separated integers N and M.
The second line contains N space-separated integers – P[1], P[2], ... P[N].

<b>Constraints:</b>
2 ≤ N ≤ 10<sup>5</sup>
1 ≤ M ≤ NPrint a single line containing N integers – the lexicographically smallest permutation achievable.Sample Input 1:
5 4
2 1 4 3 5

Sample Output 1:
1 2 3 5 4

Sample Explanation 1:
The optimal partition of the given permutation will be 2 | 1 | 4 | 3 5. The lexicographically smallest permutation of the subsegments will thus be 1 2 3 5 4.

Sample Input 2:
5 2
2 1 4 3 5

Sample Output 2:
1 4 3 5 2

Sample Explanation 2:
The optimal partition of the given permutation will be 2 | 1 4 3 5. The lexicographically smallest permutation will thus be 1 4 3 5 2., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define rep(i, n)  for(long long i=0;i<(long long)(n);i++)
#define REP(i,k,n) for(long long i=k;i<(long long)(n);i++)
#define pb emplace_back
#define lb(v,k) (lower_bound(all(v),(k))-v.begin())
#define ub(v,k) (upper_bound(all(v),(k))-v.begin())
#define fi first
#define se second
#define pi M_PI
#define PQ(T) priority_queue<T>
#define SPQ(T) priority_queue<T,vector<T>,greater<T>>
#define dame(a) {out(a);return 0;}
#define decimal cout<<fixed<<setprecision(15);
#define all(a) a.begin(),a.end()
#define rsort(a) {sort(all(a));reverse(all(a));}
#define dupli(a) {sort(all(a));a.erase(unique(all(a)),a.end());}
#define popcnt __builtin_popcountll
typedef long long ll;
typedef pair<ll,ll> P;
typedef tuple<ll,ll,ll> PP;
typedef tuple<ll,ll,ll,ll> PPP;
using vi=vector<ll>;
using vvi=vector<vi>;
using vvvi=vector<vvi>;
using vvvvi=vector<vvvi>;
using vp=vector<P>;
using vvp=vector<vp>;
using vb=vector<bool>;
using vvb=vector<vb>;
const ll inf=1001001001001001001;
const ll INF=1001001001;
const ll mod=1000000007;
const double eps=1e-10;
template<class T> bool chmin(T&a,T b){if(a>b){a=b;return true;}return false;}
template<class T> bool chmax(T&a,T b){if(a<b){a=b;return true;}return false;}
template<class T> void out(T a){cout<<a<<' ';}
template<class T> void outp(T a){cout<<'('<<a.fi<<','<<a.se<<')'<<'\n';}
template<class T> void outvp(T v){rep(i,v.size())cout<<'('<<v[i].fi<<','<<v[i].se<<')';cout<<'\n';}
template<class T> void outvvp(T v){rep(i,v.size())outvp(v[i]);}
template<class T> void outv(T v){rep(i,v.size()){if(i)cout<<' ';cout<<v[i];}cout<<'\n';}
template<class T> void outvv(T v){rep(i,v.size())outv(v[i]);}
template<class T> bool isin(T x,T l,T r){return (l)<=(x)&&(x)<=(r);}
template<class T> void yesno(T b){if(b)out("yes");else out("no");}
template<class T> void YesNo(T b){if(b)out("Yes");else out("No");}
template<class T> void YESNO(T b){if(b)out("YES");else out("NO");}
template<class T> void outset(T s){auto itr=s.begin();while(itr!=s.end()){if(itr!=s.begin())cout<<' ';cout<<*itr;itr++;}cout<<'\n';}
void outs(ll a,ll b){if(a>=inf-100)out(b);else out(a);}
ll gcd(ll a,ll b){if(b==0)return a;return gcd(b,a%b);}
ll modpow(ll a,ll b){ll res=1;a%=mod;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;}
 
int main(){
    ll n,k;cin>>n>>k;
    k--;
 
    vi v(n),rv(n);
    vb f(n+1,false);f[0]=f[n]=true;
    rep(i,n){
        cin>>v[i];v[i]--;
        rv[v[i]]=i;
    }
 
    if(k == 0)
    {
        for(int i = 0;i<n;i++)
            cout<<v[i]+1<<" ";
        return 0 ;
        
 
    }
    ll x=0;
    while(x<n){
        if(k==0)break;
        ll i=rv[x];
        if(!f[i]){
            f[i]=true;k--;
        }
        if(k==0)break;
        ll r=i;while(r+1<n&&v[r+1]==v[r]+1)r++;
        if(!f[r+1]){
            if(k==1){
                vi srt;
                rep(j,n)if(f[j]&&v[j]>=v[i])srt.pb(v[j]);
                rsort(srt);
                bool done=false;
                vb d(n+1,false);
                REP(j,v[i],n){
                    if(done)break;
                    if(!d[rv[j]]&&!f[rv[j]]){
                        f[rv[j]]=true;break;
                    }
                    if(d[rv[j]])continue;
                    assert(srt.back()==j);
                    srt.pop_back();
                    if(!srt.size())break;
                    REP(k,rv[j]+1,n){
                        if(f[k])break;
                        if(v[k]>srt.back()){
                            f[k]=true;done=true;break;
                        }
                        d[k]=true;
                    }
                }
                break;
            }
            f[r+1]=true;k--;
        }
        x=v[r]+1;
    }
    vvi al;
    rep(i,n)if(f[i]){
        vi tmp;tmp.pb(v[i]);
        REP(j,i+1,n){
            if(f[j])break;
            tmp.pb(v[j]);
        }
        al.pb(tmp);
    }
    vi srt;rep(i,al.size())srt.pb(i);
    auto cmp=[&](ll a,ll b){
        return al[a][0]<al[b][0];
    };
    sort(all(srt),cmp);
    for(ll i:srt)for(ll j:al[i])out(j+1);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not?