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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define INF ((1LL<<62)-(1LL<<31))
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))
#define bit(n,m)(((n)>>(m))&1)
int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}
ll pom(ll a,ll n,int m){ll x=1;for(a%=m;n;n/=2)n&1?x=x*a%m:0,a=a*a%m;return x;}
//#define MOD 998244353
#define MOD 1000000007
#define invp(a,p)pom(a,p-2,p)


int uniquell(ll*a,ll*ans,int cnt){
	sortup(a,cnt);
	int c=0;
	ans[0]=a[0];
	rep(i,1,cnt)if(a[i]!=ans[c])ans[++c]=a[i];
	return c+1;
}
//座圧2
//前準備：
//圧縮したい配列aに-INFとINFを追加する
//uniquell(a,comp,n)でcompに昇順の配列を作る
ll comp[200010],compcnt;
ll cocompf(ll n){
	//comp[i]<=n<comp[i+1]なるiを返す
	ll l=0,r=compcnt;
	while(r-l>1){
		ll m=(l+r)/2;
		if(comp[m]<=n)l=m;
		else r=m;
	}
	return l;
}

ll a[200010];
ll t[200010];
ll dp[100][200010];
ll pre[200010];
ll st[200010],stcnt;
ll temp[200010];
int main(){
	ll n,kk;
	scanf("%lld%lld",&n,&kk);
	rep(i,0,n)scanf("%lld",a+i),t[i]=a[i];
	compcnt=uniquell(t,comp,n);
	rep(i,0,n)pre[i]=-1;//直前にiが登場した位置
	rep(i,0,2*n){
		int j=cocompf(a[i%n]);
		if(pre[j]!=-1)dp[0][pre[j]%n]=i-pre[j]+1;
		pre[j]=i;
	}
	rep(k,1,90)rep(i,0,n){
		ll t=dp[k-1][i]+dp[k-1][(i+dp[k-1][i])%n];
		dp[k][i]=min(t,3e17);
	}
//	rep(k,0,90){rep(i,0,n)printf("%lld ",dp[k][i]);puts("");}fflush(0);
	ll crr=0;
	while(crr+dp[0][crr%n]<=n*kk){
		for(int k=88;k>=0;k--)if(crr+dp[k][crr%n]<=n*kk){
			crr+=dp[k][crr%n];
			break;
		}
	}
	if(crr!=n*kk){
		rep(i,crr%n,n){
			int j=cocompf(a[i]);
			if(temp[j]){
				while(temp[j])temp[st[--stcnt]]--;
			}else{
				st[stcnt++]=j;
				temp[j]++;
			}
		}
		rep(i,0,stcnt)printf("%lld ",comp[st[i]]);
	}
}