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<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL inf=(LL)1e18;
int n,a[200005];
vector < int > vec;
bool got[200005];
LL m;
 
int stt,lft,nx[60][200005],pr[200005];
LL ds[60][200005];
void getper(){
	int i,j,k;
	for(i=n-1;i>=0;--i) pr[a[i]]=i;
	for(i=n-1;i>=0;--i){
		nx[0][i]=pr[a[i]];
		if(nx[0][i]<=i)
			ds[0][i]=(LL)(n+nx[0][i]-i);
		else
			ds[0][i]=(LL)(nx[0][i]-i);
		++ds[0][i];
		nx[0][i]=(nx[0][i]+1)%n;
//		printf("ds %d %d = %d nx=%d\n",0,i,ds[0][i],(int)nx[0][i]);
		pr[a[i]]=i;
	}
	for(i=1;i<60;++i){
		for(j=0;j<n;++j){
			nx[i][j]=nx[i-1][nx[i-1][j]];
			ds[i][j]=ds[i-1][j]+ds[i-1][nx[i-1][j]];
			ds[i][j]=min(ds[i][j],inf);
//			if(i<3) printf("ds %d %d = %d nx=%d\n",i,j,(int)ds[i][j],nx[i][j]);
		}
	}
	stt=0;
	LL lf=m*(LL)n;
	for(i=59;i>=0;--i){
		if(lf>=ds[i][stt]){
			lf-=ds[i][stt];
			stt=nx[i][stt];
		}
	}
	assert(stt<n);
	if(lft>n+1) exit(0);
	lft=(int)lf;
}
 
int main(){
	int i,j,k;
	scanf("%d%lld",&n,&m);
	for(i=0;i<n;++i) scanf("%d",a+i);
	for(i=0;i<n;++i) if(a[i]<0 || a[i]>200000) exit(0);
	getper();
	memset(got,0,sizeof(got));
	vec.clear();
	for(i=0;i<lft;++i){
		int d=(stt+i)%n,v=a[d];
		if(!got[v]){
			vec.push_back(v);
			got[v]=1;
		}
		else{
			got[v]=0;
				if(vec.empty()) return 0;
			while(vec.back()!=v){
				got[vec.back()]=0;
				vec.pop_back();
				if(vec.empty()) return 0;
			}
			if(vec.empty()) return 0;
			vec.pop_back();
		}
	}
	for(i=0;i<(int)vec.size();++i)
		printf("%d ",vec[i]);
	printf("\n");
	return 0;
}