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>
#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)  
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define RVSORT(v) sort(v.rbegin(), v.rend());
#define ll long long
#define pb(a) push_back(a)
#define INF 999999999
#define MOD 1000000007
using namespace std;
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef pair<int, P> PP;
typedef pair<ll, LP> LPP;

int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0};

int gcd(int a,int b){return b?gcd(b,a%b):a;}

ll modpow(ll n,ll k){
  if(k == 1){
    return n;
  }else if(k == 0){
    return 1;
  }else if(k%2==0){
    ll memo = modpow(n,k/2)%MOD;
    return memo*memo%MOD;
  }else{
    return n * modpow(n,k-1)%MOD;
  }
}

int main(){

  int n,k;cin>>n>>k;
  string s,t;cin>>s;
  t = s;
  reverse(t.begin(),t.end());
  REP(i,min(k,15)){
    s = t;
    reverse(s.begin(),s.end());
    string u = s+t;
    vector<string> v;
    REP(j,s.size()){
      v.pb(u.substr(j,s.size()));
    }
    VSORT(v);
    t = v[0];
  }
  cout<<t<<endl;


 

  return 0;
}