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 <iostream>
#include <vector>
using namespace std;
const long long MAXN = 6e5+5;
long long l[64][MAXN];
long long arr[MAXN];
long long c[MAXN];
vector<long long> S;
long long p[64][MAXN];
long long n,k;
int main(){
    ios::sync_with_stdio(false);
	cin.tie(NULL);
    cin>>n>>k;
    for(long long i=0;i<n;i++){
        cin>>arr[i];
        arr[i+n] = arr[i];
    }
    for(long long i=n+n-1;i>=0;i--){
        if(i<n){
            l[0][i] = c[arr[i]]-i+1;
            p[0][i] = (c[arr[i]]+1)%n;
        }
        c[arr[i]] = i;
    }
    for(long long i=1;i<60;i++){
        for(long long j=0;j<n;j++){
            l[i][j] = min((long long)1e18,l[i-1][j]+l[i-1][p[i-1][j]]);
            p[i][j] = p[i-1][p[i-1][j]];
        }
    }
    for(long long x=0,y=0;x<n*k;){
        for(long long i=59;i>=0 && x<n*k;i--){
            if(x+l[i][y]<=n*k){
                x+=l[i][y];
                y=p[i][y];
            }
        }
        if(x==n*k){
            break;
        }
        S.push_back(arr[y]);
        x++;
        y = (y+1)%n;
    }
    for(long long x:S){
        cout<<x<<" ";
    }
}