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<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define maxn 200005
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
#define mp make_pair
#define pb push_back
#define ff first
#define ss second

int n,cnt; ll k;
int a[maxn],now[maxn],nxt[maxn],id[maxn];
pair<int,ll> b[maxn];

inline int read(){
    int x=0,f=1; char ch=getchar();
    for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
    for (;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
    return x*f;
}

bool vis[maxn];
int top,stack[maxn];

void brute_force(int x){
	for (int i=x;i<=n;i++){
		if (!vis[a[i]]) vis[a[i]]=1,stack[++top]=a[i];
		else{while (stack[top]!=a[i]) vis[stack[top]]=0,--top; --top,vis[a[i]]=0;}
	}
	for (int i=1;i<=top;i++) printf("%d%c",stack[i],i==top?'\n':' ');
}

int main(){
	n=read(); cin>>k;
	for (int i=1;i<=n;i++) a[i]=read();
	for (int i=n;i;i--) nxt[i]=now[a[i]],now[a[i]]=i;
	for (int i=n;i;i--) if (!nxt[i]) nxt[i]=now[a[i]];
	int pos=1; b[++cnt]=mp(pos,k); id[1]=1;
	while (1){
		if (nxt[pos]<=pos) --k; pos=nxt[pos]+1;
		if (k==1) break;
		if (pos>n) pos=1,--k;
		if (k==1) break; if (id[pos]) break;
		id[pos]=++cnt; b[cnt]=mp(pos,k);
	}
	if (k==1){brute_force(pos); return 0;}
	int x=b[id[pos]].ss-k; k%=x;
	for (int i=id[pos];i<=cnt;i++) if (b[id[pos]].ss-b[i].ss==k-1){brute_force(b[i].ff); return 0;}
    return 0;
}