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 INF = (1<<30);
const ll INFLL = (1ll<<60);
const ll MOD = (ll)(1e9+7);

#define l_ength size

void mul_mod(ll& a, ll b){
	a *= b;
	a %= MOD;
}

void add_mod(ll& a, ll b){
	a = (a<MOD)?a:(a-MOD);
	b = (b<MOD)?b:(b-MOD);
	a += b;
	a = (a<MOD)?a:(a-MOD);
}

int a[225816],p[410410],f[410410],s[410410]={0};
vector<int> ans;

int main(void){
	int n,i,t=0,d=0;
	ll k;
	cin >> n >> k;
	fill(f,f+n*2,n*2);
	fill(p,p+n*2,-1);
	for(i=0; i<n; ++i){
		cin >> a[i]; --a[i];
		if(p[a[i]]>=0){
			f[p[a[i]]] = i;
		}
		p[a[i]] = i;
	}
	for(i=0; i<n; ++i){
		if(p[a[i]]>=0){
			f[p[a[i]]] = i+n;
		}
		p[a[i]] = i+n;
	}
	while(true){
		if(f[t]==(n-1)){
			++d;
			break;
		}else if(f[t] == (2*n-1)){
			++d;
			s[d] = n;
			++d;
			break;
		}else if(f[t]>=n){
			++d;
			t = f[t]+1-n;
			s[d] = t;
		}else{
			t = f[t]+1;
		}
	}
	--k; k %= d;
	for(i=s[k]; i<n; ++i){
		if(f[i]<n){
			i = f[i];
			continue;
		}
		ans.push_back(a[i]+1);
	}
	for(i=0; i<ans.l_ength(); ++i){
		cout << ((i)?" ":"") << ans[i];
	}
	cout << endl;
	return 0;
}
