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<cstdio>
#include<vector>
using namespace std;

vector<int>g[200001];
int a[200001];
long long f[31][200001];
int p[31][200001];
int ans[200001];
int main(){
	int n;
	long long k;
	scanf("%d%lld",&n,&k);
	for(int i=0;i<n;i++){
		scanf("%d",&a[i]);
		g[a[i]].push_back(i);
	}
	for(int i=0;i<n;i++){
		int w=lower_bound(g[a[i]].begin(),g[a[i]].end(),i)-g[a[i]].begin();
		if(w==g[a[i]].size()-1){
			p[0][i]=(g[a[i]][0]+1)%n;
			f[0][i]=1+(g[a[i]][0]==n-1?1:0);
		}
		else{
			p[0][i]=(g[a[i]][w+1]+1)%n;
			f[0][i]=g[a[i]][w+1]==n-1?1:0;
		}
	}
	for(int i=0;i<30;i++)
		for(int j=0;j<n;j++){
			f[i+1][j]=f[i][j]+f[i][p[i][j]]+f[i][p[i][p[i][j]]]+f[i][p[i][p[i][p[i][j]]]];
			p[i+1][j]=p[i][p[i][p[i][p[i][j]]]];
		}
//	for(int i=0;i<n;i++)
//		printf("%d %d %lld\n",i,p[0][i],f[0][i]);
	int o=0;
	k--;
	for(int i=30;i>=0;i--){
		while(f[i][o]<=k){
			k-=f[i][o];
			o=p[i][o];
		}
	}
	if(k!=0)return 0;
	for(;;){
		if(f[0][o]&&(f[0][o]!=1||p[0][o])){
			ans[++ans[0]]=a[o];
			o++;
			if(o==n)break;
		}
		else{
			if(f[0][o]==1&&p[0][o]==0)break;
			o=p[0][o];
		}
	}
	for(int i=1;i<=ans[0];i++)
		printf("%d%c",ans[i],i==ans[0]?'\n':' ');
}
