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: Wrong Answer)

// #include <bits/stdc++.h>
#include <stdexcept>
#include <iostream>
#include <sstream>
#include <fstream>
#include <cassert>
#include <cstring>
#include <cstdarg>
#include <cstdio>
#include <random>
#include <cmath>
#include <ctime>
#include <functional>
#include <algorithm>
#include <complex>
#include <numeric>
#include <bitset>
#include <vector>
#include <string>
#include <queue>
#include <deque>
#include <array>
#include <list>
#include <map>
#include <set>

using namespace std;

#define PB push_back
#define MP make_pair
#define f first
#define s second
#define FOR(i,a,b) for (INT i = a; i < b; ++i)
#define RFOR(i, b, a) for (INT i = b-1; i>=a; --i)
#define FILL(a, b) memset(a, b, sizeof a)
#define ALL(a) a.begin(), a.end()
#define SZ(a) (INT)a.size()
#define INT long long 

typedef long long LL;
typedef pair<INT, INT> PII;
typedef vector<INT> VI;
typedef vector<LL> VLL;
typedef pair<LL, LL> PLL;
 
const INT INF = 1000 * 1000 * 1000;
const LL LINF = 1LL * INF * INF;
const long double PI = acos(-1.);
const double EPS = 1e-6;
LL mod = INF + 7;

const INT maxn = 200010;

VI arr[200010];
PII g[200010];
INT G[200010];

int main()
{
//  freopen("input.txt","r",stdin);
    LL n,k;
    cin >> n >> k;
    VI A(n);
    FOR(i,0,n)
    {
        INT x; cin >> x;
        arr[x].PB(i);
        A[i] = x;
    }
    if(n==1)
    {
        if(k%2)
            cout << A[0];
        return 0;
    }
    // cout << "ok" << endl;
    // return 0;
    FOR(i,0,maxn)
    {
        FOR(j,0,SZ(arr[i]))
        {
            INT to = arr[i][(j+1)%SZ(arr[i])];

            g[arr[i][j]] = MP((to+1)%n,0);
            if(SZ(arr[i])==1 || g[arr[i][j]].f <= arr[i][j])
            {
                g[arr[i][j]].s = 1;
            }
        }
    }
    // cout << "ok" << endl;
    // return 0;
    RFOR(i,n,0)
    {
        G[i] = g[i].f;
        if(g[i].s==1)
        {
            continue;
        }
        else
        {
            INT to = g[i].f;
            g[i].f = g[to].f;
            g[i].s = 1;
        }
        // cout << g[i].f << " ";
    }
    // cout << endl;
    INT lastind = 0;
    INT d = 0;
    map<INT,INT> foo1;
    set<INT> foo2;
    INT cycle;
    k--;
    while(k)
    {
        INT to = g[lastind].f;
        if(foo2.find(lastind)!=foo2.end())
        {
            cycle = d-foo1[lastind];
            k %= cycle;
            break;
        }
        else
        {
            foo1[lastind] = d;
            foo2.insert(lastind);
            lastind = to;
            d++;   
            k--;
        }
    }
    FOR(i,0,k)
        lastind = g[lastind].f;

    FOR(i,lastind,n)
    {
        if(i+1 < G[i])
        {
            i = G[i]-1;
        }
        else
        {
            if(G[i]==0 && i != n-1)
            {
                return 0;
            }
            cout << A[i] << " ";
        }
    }
    return 0 ;
}