Takahashi has a string S of length N consisting of lowercase English letters. On this string, he will perform the following operation K times:

* Let T be the string obtained by reversing S, and U be the string obtained by concatenating S and T in this order.
* Let S' be some contiguous substring of U with length N, and replace S with S'.



Among the strings that can be the string S after the K operations, find the lexicographically smallest possible one.

Constraints

* 1 \leq N \leq 5000
* 1 \leq K \leq 10^9
* |S|=N
* S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:


N K
S


Output

Print the lexicographically smallest possible string that can be the string S after the K operations.

Examples

Input

5 1
bacba


Output

aabca


Input

10 2
bbaabbbaab


Output

aaaabbaabb
#include <bits/stdc++.h>
 
using namespace std;
 
int main(void){
    int N;
    int K;
    string S;

    cin>>N;
    cin>>K;
    cin>>S;

    if(K>=N){
        int mini;
        for(int i=0;i<S.size();i++){
            if(S.at(i)<S.at(mini))
            {
                mini = i;
            }
        }

        string result = string(N,S.at(mini));
        cout << result << endl;
        return 0;
    }
    string T = S;
    reverse(T.begin(),T.end());
    string base = S+T;

    string min_S = base.substr(0,N);
    for(int i=0;i<N;i++){
        string cut_S = base.substr(i,N);
        reverse(cut_S.begin(),cut_S.end());
        if(cut_S<min_S){
            min_S = cut_S;
        }
    }

    char head = min_S.at(0);
    int times  = 1;
    while(true){
        if(head != min_S.at(times)){
            break;
        }
        times++;
    }

    string result;
    int powt = times;
    for(int i=0;i<K-1;i++){
        powt *= 2;
        if(powt>N){
        string result = string(N,min_S.at(0));
        cout << result << endl;
        return 0;
    }
    }

    K--;
    string add = min_S.substr(0,N-powt+times);
    result = string(powt-times,min_S.at(0)) + add;
    cout << result << endl;
    return 0;

}