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;

signed main(){
    //cout << setprecision(16) ;
    ios::sync_with_stdio(false);
	cin.tie(0);

    ll n,k;
    cin>>n>>k;
    ll a[n];
    vector<vector<int>> mp(200010);
    for(int i=0;i<n;i++){
        cin>>a[i];
        mp[a[i]].push_back(i);
    }
    ll cnt=0;
    ll now = 0;

    while(true){
        auto itr = upper_bound(mp[a[now]].begin(),mp[a[now]].end(),now);
        int next;
        if(itr==mp[a[now]].end()){
            itr = mp[a[now]].begin();
            cnt++;
        }
        now = *itr;
        now++;
        now%=n;
        if(now == 0) break;
    }
    
    cnt++;
    k%=cnt;
    now=0;

    while(true){
        int t=a[now];
        auto itr = upper_bound(mp[t].begin(),mp[t].end(),now);
        if(itr == mp[t].end()){
            itr = mp[t].begin();
            k--;
        }
        if(k==0) break;
        
        now = *itr;
        now++;
        now%=n;
    }
    for(;now<n;now++){
        int t=a[now];
        auto itr = upper_bound(mp[t].begin(),mp[t].end(),now);
        if(itr == mp[t].end()){
            cout<<a[now]<<" ";
        }
        else {
            now = *itr;
        }
        
    }
    cout<<endl;
}