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.

Examples

Input

3 2
1 2 3


Output

2 3


Input

5 10
1 2 3 2 3


Output

3


Input

6 1000000000000
1 1 2 2 3 3


Output




Input

11 97
3 1 4 1 5 9 2 6 5 3 5


Output

9 2 6
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=(0);i<(n);i++)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;

const int N=200005;
struct node{
	int p[N];ll c[N];
}b[65];
int a[N],las[N],fir[N],vis[N],stk[N],top,n,x;
ll K,now;

node mul(node a,node b){
	node res;
	rep(i,1,n){
		res.p[i]=b.p[a.p[i]];
		res.c[i]=b.c[a.p[i]]+a.c[i];
	}
	return res;
}

void doit(int L){
	rep(i,L,n)
		if(vis[a[i]]){
			while(stk[top]!=a[i])
				vis[stk[top--]]=0;
			vis[stk[top--]]=0;
		}
		else stk[++top]=a[i],vis[a[i]]=1;
	rep(i,1,top)printf("%d ",stk[i]);
}

int main(){
	scanf("%d%lld",&n,&K);
	rep(i,1,n){
		scanf("%d",&a[i]);
		if(!fir[a[i]])fir[a[i]]=i;
	}
	per(i,n,1){
		if(!las[a[i]]){
			b[0].p[i]=fir[a[i]]+1;
			b[0].c[i]=1;
		}
		else b[0].p[i]=las[a[i]]+1;
		if(b[0].p[i]==n+1){
			b[0].p[i]-=n;
			b[0].c[i]++;
		}
		las[a[i]]=i;
	}
	rep(i,1,60)b[i]=mul(b[i-1],b[i-1]);
	x=1,now=1;
	per(i,60,0)
		if(now+b[i].c[x]<=K){
			now+=b[i].c[x];
			x=b[i].p[x];
		}
	doit(x);
	return 0;
}