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: Accepted)

#include <iostream>
#include <iomanip>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <utility>
#include <tuple>
#include <algorithm>
#include <numeric>
#include <cstdio>
#include <cmath>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
using ll = long long;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

int main() {
  int n;
  ll k;
  cin >> n >> k;
  vector<int> a(n);
  rep(i, n) cin >> a[i];

  vector<vector<ll>> next(60, vector<ll>(n)); //ダブリング
  vector<int> prev(200010, -1);
  rep(i, 2) {
    rep(j, n) {
      if (prev[a[j]] != -1)
        next[0][prev[a[j]]%n] = n*i+j-prev[a[j]]+1; //同じ値までの距離+1
      prev[a[j]] = n*i+j; //座標
    }
  }
  for (int i=1; i<60; i++) {
    rep(j, n) {
      next[i][j] = next[i-1][j]+next[i-1][(next[i-1][j]+j)%n];
    }
  }

  ll start = 0, sum = 0;
  while (1) {
    int power = 0;
    rep(i, 60) {
      if (next[i][start]+sum >= n*k) {
        power = i-1;
        break;
      }
    }
    if (power == -1)
      break;
    sum += next[power][start];
    start = sum%n;
  }

  vector<bool> used(200010);
  deque<int> ans;
  for (int i = start; i < n; i++) { //残りの長さはn以下
    if (!used[a[i]]) {
      ans.push_back(a[i]);
      used[a[i]] = true;
    } else {
      while (1) {
        int tmp = ans.back();
        ans.pop_back();
        used[tmp] = false;
        if (tmp == a[i])
          break;
      }
    }
  }
  while (!ans.empty()) {
    printf("%d%c", ans.front(), ans.size()>1?' ':'\n');
    ans.pop_front();
  }
  return 0;
}