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: Time Limit Exceeded)

#include <bits/stdc++.h>
// #include <iostream>
using namespace std;
#define FALSE 0
#define TRUE 1

int main(){
    int64_t N;
    int64_t K;
    // int res;
    vector<int64_t> vA(200020);
    vector<int64_t> vS(400040);
    int64_t lenS = 0;

    cin >> N >> K;
    for (int i=0; i<N; i++){
        cin >> vA.at(i);
    }

    // K=2 のときの vS を愚直に作ろう
    int64_t k;
    vS.at(0) = 0;
    for (k=0; k<K; k++){
// printf("k:%lld\n", k);
        for (int i=0; i<N; i++){
            // vA.at(i) が vSに含まれるかどうか判定
// printf(" i:%d", i);
            int isS = FALSE;
            int s;
            for (s=0; s<lenS; s++){
                if (vS.at(s) == vA.at(i)){
                    isS = TRUE;
                    break;
                }
            }
// printf(" %d", isS);
            // vSに追加や削除する処理
            if (isS == FALSE) {
                // isS == FALSE
                vS.at(lenS) = vA.at(i);
                lenS++;
            }
            else {
                // isS == TRUE
                lenS = s;
            }
// printf(" %lld\n", lenS);
        }
        if (lenS == 0) break;
    }

    // lenS が 0となったタイミングでbreakして、
    // 処理回数のあまりをとる処理
    int64_t modK;
    modK = (lenS==0)? (K % (k+1)): K;
// printf("\nmodK:%lld\n", modK);

    // modK回だけ再度処理する
    lenS = 0;
    vS.at(0) = 0;
    for (k=0; k<modK; k++){
// printf("k:%lld\n", k);
        for (int i=0; i<N; i++){
// printf(" i:%d", i);
            // vA.at(i) が vSに含まれるかどうか判定
            int isS = FALSE;
            int s;
            for (s=0; s<lenS; s++){
                if (vS.at(s) == vA.at(i)){
                    isS = TRUE;
                    break;
                }
            }
// printf(" %d", isS);
            // vSに追加や削除する処理
            if (!isS) {
                vS.at(lenS) = vA.at(i);
                lenS++;
            }
            else {
                lenS = s;
            }
// printf(" %lld\n", lenS);
        }
    }


    for (int64_t i=0; i<lenS; i++)
        cout << vS.at(i) << " "; 
    cout << endl;
    return 0;
}