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>
using namespace std;
typedef long long LL;
const int N=200200;
int n;
int a[N];
int na[N];
int nxt[N];
LL k;
bool in[N];
int s[N];
int st=0;
int main(){
	scanf("%d",&n);
	cin>>k;
	for(int i=0;i<n;i++){
		scanf("%d",&a[i]);
	}
	for(int i=n-1;i>=0;i--){
		na[a[i]]=n+i;
	}
	for(int i=n-1;i>=0;i--){
		nxt[i]=na[a[i]];
		// cerr<<i<<" "<<nxt[i]<<endl;
		na[a[i]]=i;
	}
	int t=0;
	for(int i=0;;){
		// cerr<<i<<endl;
		i=nxt[i]+1;
		while(i>=n){
			i-=n;
			t++;
		}
		if(i==0){
			break;
		}
	}
	// cerr<<"t = "<<t<<endl;
	k%=t;
	int i=0;
	for(;k>5;){
		i=nxt[i]+1;
		while(i>=n){
			i-=n;
			k--;
		}
	}
	while(k>0){
		if(in[a[i]]){
			while(in[a[i]]){
				in[s[--st]]=false;
			}
		}
		else{
			in[a[i]]=true;
			s[st++]=a[i];
		}
		i++;
		if(i==n){
			i=0;
			k--;
		}
	}
	for(int i=0;i<st;i++){
		printf("%d ",s[i]);
	}
	printf("\n");
	return 0;
}