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>
#define be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007, INF=mod*mod*3LL;
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;

int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    ll n, k;
    cin >> n >> k;
    vector<ll> a(n);
    for(int i=0;i<n;i++) cin >> a[i];
    vector<ll> b(200001);
    for(int i=n-1;i>=0;i--) b[a[i]] = i;
    vector<ll> c(n), inc(n);
    for(int i=n-1;i>=0;i--){
    	if(b[a[i]] <= i) inc[i] = 1;
    	else  inc[i] = 0;
    	c[i] = b[a[i]];
    	b[a[i]] = i;
    }
    ll num = 1, now = 0;
    while(1){
    	num += inc[now];
    	now = (c[now] + 1) % n;
    	if(now == 0) break;
    }
    k %= num;
    if(k == 0)return 0;

    num = 1;
    while(!(num == k && inc[now])){
    	num += inc[now];
    	now = c[now] + 1;
    }
    for(int i=now;i<n;i++){
    	if(inc[i] == 0) i = c[i];
    	else cout << a[i] << " ";
    }
    cout << endl;
    return 0;
}
