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<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#define ll long long
#define maxn 200010
inline ll read()
{
	ll x=0; char c=getchar(),f=1;
	for(;c<'0'||'9'<c;c=getchar())if(c=='-')f=-1;
	for(;'0'<=c&&c<='9';c=getchar())x=x*10+c-'0';
	return x*f;
}
inline void write(ll x)
{
	static int buf[20],len; len=0;
	if(x<0)x=-x,putchar('-');
	for(;x;x/=10)buf[len++]=x%10;
	if(!len)putchar('0');
	else while(len)putchar(buf[--len]+'0');
}
inline void writeln(ll x){write(x); putchar('\n');}
inline void writesp(ll x){write(x); putchar(' ');}
int a[maxn],nxt[maxn],pos[maxn],dep[maxn];
ll n,k;
void print(int s)
{
	static int mark[maxn],ans[maxn],tot;
	memset(mark,0,sizeof(mark));
	tot=0;
	for(int i=s;i<=n;i++)
		if(!mark[a[i]])mark[a[i]]=1,ans[++tot]=a[i];
		else{
			while(ans[tot]!=a[i])mark[ans[tot--]]=0;
			mark[ans[tot--]]=0;
		}
	for(int i=1;i<=tot;i++)
		writesp(ans[i]);
	putchar('\n');
}
int main()
{
	n=read(); k=read();
	for(int i=1;i<=n;i++)
		a[i]=read();
	for(int i=n;i;i--)
		pos[a[i]]=i;
	for(int i=n;i;i--){
		nxt[i]=pos[a[i]];
		pos[a[i]]=i;
	}
	// for(int i=1;i<=n;i++)
	// 	printf("%d %d %d\n",i,a[i],nxt[i]);
	int now=nxt[1]%n+1,len=(nxt[1]<=1)+(nxt[1]==n);
	while(now!=1){
		// puts("QAQ");
		len+=(nxt[now]<=now)+(nxt[now]==n);
		now=nxt[now]%n+1;
	}
	// writeln(len);
	k=(k-1)%len;
	now=nxt[1]%n+1; len=(nxt[1]<=1);
	if(len>k){
		print(now);
		return 0;
	}
	len+=(nxt[1]==n);
	if(len>k){
		print(n+1);
		return 0;
	}
	while(now!=1){
		// puts("TAT");
		len+=(nxt[now]<=now);
		if(len>k){
			print(now);
			return 0;
		}
		len+=(nxt[now]==n);
		if(len>k){
			print(n+1);
		}
		now=nxt[now]%n+1;
	}
	return 0;
}