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: Accepted)

#include <iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#define ll long long
using namespace std;

ll n,k;
int a[300000];
int t=1;
int s[300000];
int flag[300000];

void work()
{
    for (int i=1; i<=k; ++i)
    {
        for (int j=1; j<=n; ++j)
            {
                if(flag[a[j]]==0)
                {
                    flag[a[j]]++;
                    t+=1;
                    s[t]=a[j];
                }else
                {
                    while(s[t]!=a[j]) {flag[s[t]]--,t--;}
                    t--;
                    flag[a[j]]--;
                }
            }
        cout<<i<<endl;
        for (int j=1; j<=t; ++j)
            cout<<s[j]<<' ';
        cout<<endl;
        cout<<endl;
    }
}

int firstifi[300000]; // if begin with i, the number will be firstifi[i]
int nexti[300000]; // nexti[i] = j where min(a[i]==a[j])
int loc[300000]; //loc[a[i]] = j where min(a[i]==a[j])
int firstloc[300000]; //firstloc[num] = loc

void work2()
{
    for (int i=n; i>=1; --i)
    {
        firstloc[a[i]]=i;
        if (loc[a[i]]==0)
            nexti[i] = 0;
        else
            nexti[i] = loc[a[i]];
        loc[a[i]] = i;
    }

    //for (int i=1; i<=n; ++i)
    //    cout<<nexti[i]<<' ';
    //sssssssssssssssssssssssscout<<endl;

    for (int i=n; i>=1; --i)
        if (nexti[i]==0)
            firstifi[i] = a[i];
        else
            firstifi[i] = firstifi[nexti[i]+1];

    //for (int i=1; i<=n; ++i)
    //    cout<<firstifi[i]<<' ';
    //cout<<endl;
    int nownum = 0;
    for (int i=1; i<=k; ++i)
    {
        if (nownum==0)
            nownum = firstifi[1];
        else
            nownum = firstifi[firstloc[nownum]+1];
        s[t] = nownum;
        t++;
        if (nownum==0) break;
    }
    s[0]=0;
    t--;
    //for (int i=1; i<=t; ++i)
    //    cout<<s[i]<<' ';
    //cout<<endl;

    k = k-1;
    k = k%t;

    nownum = s[k];

    if (nownum!=0)
    {
            s[1]=nownum;
            t=1;
            flag[nownum]=1;
    }else
    {
        t=0;
    }


        for (int j=1; j<=n; ++j)
            {
                if(flag[a[j]]==0)
                {
                    flag[a[j]]++;
                    t+=1;
                    s[t]=a[j];
                }else
                {
                    while(s[t]!=a[j]) {flag[s[t]]--,t--;}
                    t--;
                    flag[a[j]]--;
                }
            }
        if (t==0)
        {
            cout<<endl;
            return;
        }

        for (int j=1; j<=t; ++j)
            cout<<s[j]<<' ';
        cout<<endl;


}
int main()
{
    cin>>n>>k;
    for (int i=1; i<=n; ++i)
        scanf("%d",&a[i]);
    work2();
    return 0;
}