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)

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<cmath>

using namespace std;

int main(int argc, char const *argv[]){
    long long n,k, l;
    vector<long long> a, b(20000,-1), c(20000,-1);
    cin >> n >> k;
    for(long long i=0;i<n;i++){
        long long _a;
        cin >> _a;
        a.push_back(_a);
    }

    int flag=0;

    l=0;
    for(int m=0;m<k;m++){
        for(int i=0;i<n;i++){
            if(b[a[i]]<0){
                c[l]=a[i];
                b[a[i]] = l;
                l++;
            }else{
                l = b[a[i]];
                for(int j=b[a[i]];j<n;j++){
                    if(c[j]==(-1)){
                        break;
                    }else{
                        b[c[j]] = -1;
                        c[j] = -1;
                    }
                }
            }
        }
        if(c[0]==(-1)){
            k = k % (m+1);
            flag = 1;
            break;
        }
    }

    if(flag==1){
        for(int m=0;m<k;m++){
            for(int i=0;i<n;i++){
                if(b[a[i]]<0){
                    c[l]=a[i];
                    b[a[i]] = l;
                    l++;
                }else{
                    l = b[a[i]];
                    for(int j=b[a[i]];j<n;j++){
                        if(c[j]==(-1)){
                            break;
                        }else{
                            b[c[j]] = -1;
                            c[j] = -1;
                        }
                    }
                }
            }
        }
    }

    for(auto v:c){
        if(v==(-1)) break;
        cout << v << " ";
    }
    cout << endl;
    
    return 0;
}