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;
const int maxn=1e4+10;
char s[maxn],res[maxn];
char c;
int n,k,len[maxn],mx;
bool check(int p)
{
    for(int i=1;i<=n;i++)
    if(s[i+p-1]>res[i])return false;
    else if(s[i+p-1]<res[i])return true;
    return false;
}
int main()
{
    c='z'+1;
    scanf("%d%d",&n,&k);
    scanf("%s",s+1);
    for(int i=1;i<=n;i++)c=min(c,s[i]),s[n*2-i+1]=s[i];
    for(int i=1;i<=n;i++)res[i]='z'+1;
    if(k>=15)
    {
        for(int i=1;i<=n;i++)printf("%c",c);
        printf("\n");
        return 0;
    }
    if(k==1)
    {
        for(int i=1;i<=n+1;i++)
        if(check(i))
        {
            for(int j=1;j<=n;j++)res[j]=s[i+j-1];
        }           
        printf("%s",res+1);
        return 0;
    }
    for(int i=1;i<=2*n;i++)len[i]=s[i]==c?len[i-1]+1:0,mx=max(mx,len[i]);
    k--;
    if((mx<<k)>=n)
    {
        for(int i=1;i<=n;i++)printf("%c",c);
        printf("\n");
        return 0;
    }
    int now=mx<<k;
    for(int i=1;i<=n+mx;i++)
    if(len[i]==mx&&check(i+1))
    for(int j=1;j<=n;j++)res[j]=s[i+j];
    for(int i=1;i<=now;i++)printf("%c",c);
    for(int i=now+1;i<=n;i++)printf("%c",res[i-now]);
    printf("\n");
}
