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>
using namespace std;
#define pp pair<int,int>
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define ll long long
ll MOD=1000000007;
int inf=1000000000;
ll INF=10000000000000000;
int main(){
    ll n;
    ll k;
    cin >> n >> k;
    vector<ll> a(n),c(n,inf),d(n,0);
    vector<vector<int>> b(200010,vector<int>(0));
    rep(i,n){
        int g;
        cin >> g;
        a.at(i)=g;
        b.at(g).push_back(i);
    }
    //return 0;
    ll u=0,t=0;
    while (true){
        int tt=t;
        if (c.at(t)!=inf) break;
        d.at(t)=u;
        auto it=upper_bound(b.at(a.at(t)).begin(),b.at(a.at(t)).end(),t);
        if (it==b.at(a.at(t)).end()) {t=b.at(a.at(t)).at(0)+1;u++;}
        else t=*it+1;
        if (t==n) {t=0;u++;}
        c.at(tt)=t;
    }
    k=(k-d.at(t))%u;
    int y;
    if (k==0) {cout << endl;return 0;}
    while (true){
        y=t;
        if (t>=c.at(t) || t+1==c.at(t)) k--;
        if (k==0) break;
        t=c.at(t);
    }
    vector<int> r(0);
    while (true){
        
        //cout << y << " " << c.at(y) << endl;
        auto it=upper_bound(b.at(a.at(y)).begin(),b.at(a.at(y)).end(),y);
        if (it==b.at(a.at(y)).end()) {r.push_back(a.at(y));y++;}
        else y=c.at(y);
        //cout << y << endl;
        if (y>=n || y==0) break;
    }
    if (r.size()==0) {cout << endl;return 0;}
    rep(i,r.size()-1) cout << r.at(i) << " ";
    cout << r.at(r.size()-1) << endl;
}