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;
#define ft first
#define sd second
#define fr(i,n) for(int i=0;i<(n);++i)
#define Fr(i,n) for(int i=1;i<=(n);++i)
#define ifr(i,n) for(int i=(n)-1;i>=0;--i)
#define iFr(i,n) for(int i=(n);i>0;--i)

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    ll n,k,t=0,w=0,s=0;
    cin>>n>>k;
    vector<ll> a(n),nx(n);
    vector<vector<ll>> b(200001);
    fr(i,n) cin>>a[i],b[a[i]].emplace_back(i);
    Fr(i,200000) fr(j,b[i].size()) nx[b[i][j]]=b[i][(j+1)%b[i].size()];
    while(1){
        if(w==n) break;
        t+=nx[w]+1-w;
        if(nx[w]<=w) t+=n;
        w=nx[w]+1;
    }
    t/=n;
    k%=t;
    w=0;
    while(1){
        if(s+nx[w]+1-w+(nx[w]<=w)*n>k*n) break;
        s+=nx[w]+1-w;
        if(nx[w]<=w) s+=n;
        w=nx[w]+1;
    }
    if(s==k*n) return 0;
    for(;w<n;++w){
        if(nx[w]<=w) cout<<a[w]<<" ";
        else w=nx[w];
    }/*
    vector<ll> ans;
    vector<bool> vis(200001);
    for(;w<n;++w){
        if(vis[a[w]]){
            while(*ans.rbegin()!=a[w]) ans.pop_back();
            ans.pop_back();
        }
        else ans.emplace_back(a[w]),vis[a[w]]=1;
    }
    for(auto i:ans) cout<<i<<" ";*/
    cout<<endl;
}