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: Wrong Answer)

#include<iostream>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>
#include<algorithm>
#include<cstring>
#include<string>
#include<cassert>
#include<cmath>
#include<climits>
#include<iomanip>
#include<bitset>
#include<unordered_map>

using namespace std;

#define rep(i,m,n) for(int (i)=(int)(m);(i)<(int)(n);(i)++)
#define rep2(i,m,n) for(int (i)=(int)(n)-1;(i)>=(int)(m);(i)--)
#define REP(i,n) rep(i,0,n)
#define FOR(i,c) for(decltype((c).begin())i=(c).begin();i!=(c).end();++i)
#define ll long long
#define ull unsigned long long
#define all(hoge) (hoge).begin(),(hoge).end()
#define en '\n'
typedef pair<ll, ll> P;
constexpr long long INF = 1LL << 60;
constexpr int INF_INT = 1 << 25;
constexpr long long MOD = (ll)1e9 + 7;
typedef vector<ll> Array;
typedef vector<Array> Matrix;
template <class T> using vec = vector<T>;
template <class T> using vvec = vector<vec<T>>;


template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}

struct Edge {
	int to, cap, rev;
	Edge(int _to, int _cap, int _rev) {
		to = _to; cap = _cap; rev = _rev;
	}
};
typedef vector<Edge> Edges;
typedef vector<Edges> Graph;

void add_edge(Graph& G, int from, int to, int cap, bool revFlag, int revCap) {
	G[from].push_back(Edge(to, cap, G[to].size()));
	if (revFlag)G[to].push_back(Edge(from, revCap, G[from].size() - 1));
}



void solve() {
	ll n, k;
	cin >> n >> k;
	Array a(n);
	REP(i, n) cin >> a[i];

	Array next(n,n+1);//添字iが消えた後の位置
	Array first(202020, n);
	Array pre(202020, n);
	REP(i, n) {
		if (first[a[i]] == n) {
			first[a[i]] = i;
			pre[a[i]] = i;
			continue;
		}

		next[pre[a[i]]] = i - pre[a[i]] + 1;
		next[i] = n + first[a[i]] - i + 1;
		pre[a[i]] = i;
	}

	Matrix t(41,Array(n,-1));//2^i周期進んだ時の位置
	REP(i, n) {
		if (t[0][i] != -1) continue;
		ll now = i;
		Array p;
		p.push_back(now);
		while (now < n) {
			now += next[now];
			if (now < n) {
				if (t[0][now] != -1) {
					now = t[0][now];
					break;
				}
				p.push_back(now);
			}
		}
		for(auto j:p) t[0][j] = now % n;//1周期進んだ位置
	}

	REP(i, 40) {
		REP(j, n) {
			t[i + 1][j] = t[i][t[i][j]];//ダブリング
		}
	}

	ll pos = 0;
	REP(i, 41) {
		if ((k - 1) & (1LL << i)) {
			pos = t[i][pos];
		}
	}

	while (pos != n) {
		while (pos < n && pos + next[pos] <= n) {
			pos += next[pos];
		}
		if (pos < n) {
			cout << a[pos] << " ";
			pos++;
		}
	}
	cout << en;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	solve();
	//int t; cin >> t; REP(i, t) solve();

	return 0;
}