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;
    typedef pair<int,int>pii;
    string s,ss,s2,t,ans,ff;
    vector<pii>v;
    int n,k,i,z,l;
    char mn=127;
    int main(){
    	ios::sync_with_stdio(0);cin.tie(0);
    	cin>>n>>k>>s;ss=s;reverse(ss.begin(),ss.end());ss=s+ss;
    	for(char c:s)mn=min(mn,c);
    	for(i=0;i<n;++i)if(s[i]==mn)v.push_back(pii(++z,i));else z=0;
    	sort(v.begin(),v.end());
    	if(k>15 || (v.back().first<<k-1)>=n){for(i=0;i<n;++i)cout<<mn;return 0;}
    	if(z){
    		if((z<<k)>=n){for(i=0;i<n;++i)cout<<mn;return 0;}
    		l=n-(z<<k);
    		ff=string(n-l,mn)+ss.substr(n+z,l);
    	}else ff=s;
    	l=n-(v.back().first<<k-1);
    	for(pii u:v)if(u.first==v.back().first){
    		t=ss.substr(n*2-1-u.second-l,l);reverse(t.begin(),t.end());
    		if(s2.empty() || t<s2)s2=t;
    	}
    	ans=min(string(n-l,mn)+s2,min(s,ff));
    	cout<<ans<<endl;
    	return 0;
    }