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 mn=2e5+10;
ll a[mn],nxt[mn],cur[mn],sp[mn][60];
bool in[mn];
ll inf=0x3f3f3f3f3f3f3f3f;
int main()
{
    ll n,k,i,j;
    scanf("%lld%lld",&n,&k);
    memset(cur,-1,sizeof(nxt));
    for(i=0;i<n;i++)scanf("%lld",a+i);
    for(i=n-1;i>=0;i--){
        nxt[i]=cur[a[i]];
        cur[a[i]]=i;
    }
    for(i=n-1;i>=0;i--){
        if(nxt[i]==-1)nxt[i]=cur[a[i]]+n;
    }
    for(i=0;i<n;i++){
        sp[i][0]=nxt[i]-i+1;
    }
    for(i=1;i<60;i++)for(j=0;j<n;j++)sp[j][i]=min(inf,sp[j][i-1]+sp[(j+sp[j][i-1])%n][i-1]);
    ll cum=0;
    for(i=59;i>=0;i--){
        if(cum+sp[cum%n][i]<=n*k)cum+=sp[cum%n][i];
    }
    vector<int>ans;
    for(i=cum;i<n*k;i++){
        if(!in[a[i%n]])ans.push_back(a[i%n]),in[a[i%n]]=1;
        else while(in[a[i%n]]){
            in[ans[ans.size()-1]]=0;
            ans.pop_back();
        }
    }
    for(int x:ans)printf("%d ",x);
}
