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;
long long a[2000010];
long long n;
long long k;
long long s[2000010];
long long m=0;
long long ne[3000000];
long long p[3000000];
void solve(long long id) {
    //cout<<id<<endl;
    while(id<n && p[id]<=id) {
        printf("%lld ",a[id]);
        id++;
    }
    if(id==n || p[id]+1==n) return;
    solve(p[id]+1);
}
int main() {
    scanf("%lld %lld",&n,&k);
    for(long long i=0;i<n;i++) scanf("%lld",&a[i]);
    for(long long i=n-1;i>=0;i--) {
        ne[a[i]]=i;
    }
    for(long long i=n-1;i>=0;i--) {
        p[i]=ne[a[i]];
        ne[a[i]]=i;
        //cout<<i<<" "<<p[i]<<endl;
    }
    long long it=0;
    long long cur=0;
    while(cur+1<k && (it!=0 || cur==0)) {
        //cout<<it<<" "<<cur<<endl;
        if(p[it]<=it) cur++;
        if(p[it]+1==n) cur++;
        it=p[it]+1;
        if(it==n) it=0;
    }
    if(cur+1==k) {
        solve(it);
    }
    else {
        k=k%((long long)cur);
        it=0;
        cur=0;
        if(k==0) return 0;
        while(cur+1<k) {
            if(p[it]<=it || p[it]+1==n) cur++;
            it=p[it]+1;
            if(it==n) it=0;
        }
        solve(it);
    }
    return 0;
}
