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>
#define LL long long
#define fr(i,x,y) for(int i=(x);i<=(y);i++)
#define rf(i,x,y) for(int i=(x);i>=(y);i--)
#define frl(i,x,y) for(int i=(x);i<(y);i++)
using namespace std;
const int N=200002;
int n,a[N];
int b[N];
int h[N],w;
int to[N];
LL K;

void read(int &x){ scanf("%d",&x); }
void read(LL &x){ scanf("%lld",&x); }

void gooo(int p,int s){
	memset(b,0,sizeof b);
	fr(i,1,s){
		if (!b[a[p]]) h[++w]=a[p],b[a[p]]=1;
	 	 else{
	 	 	while(h[w]!=a[p]) b[h[w]]=0,w--;
	 	 	b[h[w]]=0;w--;
		  }
		p=p%n+1;
	}
	fr(i,1,w) printf("%d ",h[i]);
}

void goo(int p,LL s){
	while(to[p]-p+1<=s){
		s-=to[p]-p+1;
		p=to[p]%n+1;
	}
	gooo(p,s);
}

int main(){
	read(n);read(K);
	fr(i,1,n) read(a[i]);
	rf(i,n,1) b[a[i]]=i+n;
	rf(i,n,1) to[i]=b[a[i]],b[a[i]]=i;
	memset(b,0,sizeof b);
	int p=1;LL s=0,sc=0;
	while(!b[p]){
		b[p]=1;
		s+=to[p]-p+1;
		p=to[p]%n+1;
	}
	int q=p;
	while(1){
		sc+=to[q]-q+1;
		q=to[q]%n+1;
		if (q==p) break;
	}
	if (n*K<=s) goo(1,n*K);
	 else goo(p,(n*K-s)%sc);
	return 0;
}