#include<cstdio>
#include<cassert>
#include<vector>
#include<iostream>
#include<string>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<functional>
#include<utility>
#include<cstring>
#include<numeric>
#include<algorithm>
#include<atcoder/math>
#include<atcoder/modint>
#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
using namespace atcoder;
using ll = long long;
using ull = unsigned long long;
using Mint = modint998244353;
using mint = modint;
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define rrep(i, n) for (int i = (int)(n)-1; i >= 0; --i)
#define rep2(i, a, b) for (int i = (int)a; i < (int)(b); ++i)
#define rrep2(i, a, b) for (int i = (int)(b)-1; i >= (int)(a); --i)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
constexpr int dx[] = {-1,0,1,0};
constexpr int dy[] = {0,-1,0,1};
constexpr int MAX_N = 100000;
//ダイクストラ法：makedijkstra
//エラトステネスの篩：makesieve
//テストケースが複数の場合：multest

int main(){
    int N, K;
    scanf("%d %d", &N, &K);
    int A[N];
    for(int i = 0; i < N; ++i) scanf("%d", A + i);
    int last = A[0], l = 0;
    int diff_idx[N];
    bool is_bigger_than_next[N];
    rep(i, N-1){
        if(A[i+1] != A[i]){
            rep2(j, l, i+1){
                diff_idx[j] = i+1;
                is_bigger_than_next[j] = A[i] > A[i+1];
            }
            l = i+1;
        }
    }
    rep2(i, l, N){
        diff_idx[i] = N;
        is_bigger_than_next[i] = false;
    }
    int answer[N];
    iota(answer, answer + N, 0);
    auto comp = [&](int a, int b){
        assert(a < b);
        if(diff_idx[a] > b) return false;
        return is_bigger_than_next[a];
    };
    nth_element(answer, answer + K - 1, answer + N, [&](int a, int b){
        return a < b ? comp(a, b) : !comp(b, a);
    });
    rep(i, answer[K-1]){
        printf("%d", A[i]);
        printf(i < N-1 ? " " : "\n");
    }
    rep2(i, answer[K-1]+1, N){
        printf("%d", A[i]);
        printf(i < N-1 ? " " : "\n");
    }
}