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 <cstdio>
#include <iostream>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <stack>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

int main() {
    ll N, K;
    cin >> N >> K;

    auto A = vector<ll>(N);
    for(auto &a : A)
        cin >> a;

    auto s = stack<ll>();
    auto pos = vector<ll>(200000);
    for(auto &p : pos)
        p = -1;

    for(ll k=0; k<K; k++) {
        for(ll i=0; i<N; i++) {
            auto a = A[i];
            if(pos[a] == -1) {
                s.push(a);
                pos[a] = k * N + i;
                continue;
            }

            while(s.top() != a) {
                pos[s.top()] = -1;
                s.pop();
            }
            pos[a] = -1;
            s.pop();
        }
    }

    auto ans = vector<ll>(s.size());
    for(ll i=s.size()-1; i>=0; i--) {
        ans[i] = s.top();
        s.pop();
    }

    for(auto &x : ans)
        cout << x << ' ';
    cout << endl;

    return 0;
}