Problem:
Score : 700 points

Problem Statement

We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}).
Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.

Snuke has an integer sequence s, which is initially empty.
For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:

If s does not contain X_i: add X_i to the end of s.

If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.

Find the elements of s after Snuke finished the operations.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq K \leq 10^{12}

1 \leq A_i \leq 2 \times 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A_0 A_1 \cdots A_{N-1}

Output

Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.

Sample Input 1

3 2
1 2 3

Sample Output 1

2 3

In this case, X=(1,2,3,1,2,3).
We will perform the operations as follows:

i=0: s does not contain 1, so we add 1 to the end of s, resulting in s=(1).

i=1: s does not contain 2, so we add 2 to the end of s, resulting in s=(1,2).

i=2: s does not contain 3, so we add 3 to the end of s, resulting in s=(1,2,3).

i=3: s does contain 1, so we repeatedly delete the element at the end of s as long as s contains 1, which causes the following changes to s: (1,2,3)→(1,2)→(1)→().

i=4: s does not contain 2, so we add 2 to the end of s, resulting in s=(2).

i=5: s does not contain 3, so we add 3 to the end of s, resulting in s=(2,3).

Sample Input 2

5 10
1 2 3 2 3

Sample Output 2

3

Sample Input 3

6 1000000000000
1 1 2 2 3 3

Sample Output 3

s may be empty in the end.

Sample Input 4

11 97
3 1 4 1 5 9 2 6 5 3 5

Sample Output 4

9 2 6

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

///Bismillahir Rahmanir Rahim
#include<bits/stdc++.h>

#define                           int                         long long
#define                           fi                          first
#define                           si                          second
#define                           mp                          make_pair
#define                           pb                          push_back
#define                           pi                          pair<int,int>
#define                           clr(x)                      memset(x,0,sizeof(x));
#define                           f(i,l,r)                    for(int i=l;i<=r;i++)
#define                           rf(i,r,l)                   for(int i=r;i>=l;i--)
#define                           done(i)                     cout<<"done = "<<i<<endl;
#define                           fast                        ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);




using namespace std;

const int inf=1e18;
const int mod=1e9+7;
const int M=200009;

inline int bigmod(int B,int P){int R=1;while(P>0){if(P&1){R=(R*B)%mod;}P>>=1;B=(B*B)%mod;}return R;}
inline int ad(int x,int y){int ret=(x%mod+y%mod)%mod;if(ret<0){ret+=mod,ret=ret%mod;}return ret;}
inline int sub(int x,int y){int ret=((x%mod)-(y%mod)+mod)%mod;if(ret<0){ret+=mod,ret=ret%mod;}return ret;}
inline int gun(int x,int y){int ret=((x%mod)*(y%mod))%mod;if(ret<0){ret+=mod,ret=ret%mod;}return ret;}

int a[M+4];
 main()

{
    fast
    int n,k;
    cin>>n>>k;
     f(i,1,n)cin>>a[i];
    int dp[n+4],nxt[n+4],last[n+4];
    int first_pos[n+4];
    memset(dp,-1,sizeof dp);
    memset(nxt,-1,sizeof nxt);
    memset(last,-1,sizeof last);
    memset(first_pos,-1,sizeof first_pos);
    rf(i,n,1)
    {
       int x=a[i];
       if(last[x]!=(-1))
       {
           nxt[i]=last[x]+1;
       }
       last[x]=i;
       first_pos[x]=i;
    }
    dp[n]=a[n];
    dp[n+1]=0;
    rf(i,n-1,1)
    {
      if(nxt[i]==-1)
      {
          dp[i]=a[i];
      }
      else
      {
          dp[i]=dp[nxt[i]];
      }
    }

    int tmp=(k-1)%n;
    first_pos[0]=0;
    int yo=0;
    int p=0;
    while(1)
    {
        int st=first_pos[yo]+1;
        yo=dp[st];
        p++;
        if(yo==0)
        {
            break;
        }
    }
    tmp=(k-1)%p;
    yo=0;
      f(i,0,tmp-1)
    {
        int st=first_pos[yo]+1;
        yo=dp[st];
    }
    //cout<<yo<<" # "<<endl;
    int now=first_pos[yo]+1;
    int nys[n+2];
    vector<int>ses;
    memset(nys,-1,sizeof nys);
    for(int i=now;i<=n;i++)
    {
          int x=a[i];
          if(nys[x]==-1)
          {
              ses.pb(x);
              nys[x]=1;
          }
          else
          {

             while(ses.size())
             {
                 int val=ses.back();
                 ses.pop_back();
                 nys[val]=-1;
                 if(val==x)break;
             }

          }
    }
//    f(i,1,n)
//    {
//        cout<<dp[i]<<" ";
//    }
//    cout<<endl;
//    cout<<"now "<<now<<endl;
//   // f(i,1,n)cout<<a[i]<<" ";cout<<endl;
//    cout<<"....."<<endl;

  // cout<<k<<" ::: ";
    for(auto x:ses)
    {
        cout<<x<<" ";
    }
    cout<<endl;

    return 0;

}