#include <bits/stdc++.h>

#pragma GCC target("avx2")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
//#include <boost/multiprecision/cpp_int.hpp>
using namespace std;

//using namespace boost::multiprecision;
//#include<atcoder/all>
//#include<atcoder/segtree>
//#include <atcoder/scc>
//#include <atcoder/dsu>
//using namespace atcoder;

using dou =long double;
string yes="yes";
string Yes="Yes";
string YES="YES";
string no="no";
string No="No";
string NO="NO";

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

//ll mod = 998244353ll;
ll mod = 1000000007ll;
//ll mod;
//const ll mod = 4;


struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime modhttps://atcoder.jp/contests/abc166/submit?taskScreenName=abc166_f
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};

istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}


#define rep(i, n)         for(ll i = 0; i < (ll)(n); i++)
//#define rep(i, n)         for(int i = 0; i < (int)(n); i++)
#define brep(n)           for(int bit=0;bit<(1<<n);bit++)
#define bbrep(n)           for(int bbit=0;bbit<(1<<n);bbit++)
#define erep(i,container) for (auto &i : container)
#define itrep(i,container) for (auto i : container)
#define irep(i, n)        for(ll i = n-1; i >= (ll)0ll; i--)
#define rrep(i,m,n) for(ll i = m; i < (ll)(n); i++)
#define reprep(i,j,h,w) rep(i,h)rep(j,w)
#define repreprep(i,j,k,h,w,n) rep(i,h)rep(j,w)rep(k,n)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define VEC(type,name,n) std::vector<type> name(n);rep(i,n)std::cin >> name[i];
#define pb push_back
#define pf push_front
#define query int qq;std::cin >> qq;rep(qqq,qq)
#define lb lower_bound
#define ub upper_bound
#define fi first
#define se second
#define itn int
#define mp make_pair
//#define sum(a) accumulate(all(a),0ll)
#define keta fixed<<setprecision
#define kout(d) std::cout << keta(10)<< d<< std::endl;
#define vout(a) erep(qxqxqx,a)std::cout << qxqxqx << ' ';std::cout  << std::endl;
#define vvector(name,typ,m,n,a)vector<vector<typ> > name(m,vector<typ> (n,a))
//#define vvector(name,typ,m,n)vector<vector<typ> > name(m,vector<typ> (n))
#define vvvector(name,t,l,m,n,a) vector<vector<vector<t> > > name(l, vector<vector<t> >(m, vector<t>(n,a)));
#define vvvvector(name,t,k,l,m,n,a) vector<vector<vector<vector<t> > > > name(k,vector<vector<vector<t> > >(l, vector<vector<t> >(m, vector<t>(n,a)) ));
//#define case std::cout <<"Case #" <<qqq+1<<":"
#define RES(a,iq,jq) a.resize(iq);rep(iii,iq)a[iii].resize(jq);

#define RESRES(a,i,j,k) a.resize(i);rep(ii,i)a[ii].resize(j);reprep(ii,jj,i,j){a[ii][jj].resize(k);};
#define res resize
#define as assign
#define ffor for(;;)
#define ppri(a,b) std::cout << a<<" "<<b << std::endl
#define pppri(a,b,c) std::cout << a<<" "<<b <<" "<< c<<std::endl
#define ppppri(a,b,c,d) std::cout << a<<" "<<b <<" "<< c<<' '<<d<<std::endl
#define aall(x,n) (x).begin(),(x).begin()+(n)
#define SUMI(a) accumulate(all(a),0) 
#define SUM(a) accumulate(all(a),0ll) 
#define stirng string
#define gin(a,b) int a,b;std::cin >> a>>b;a--;b--;
#define popcount __builtin_popcount
#define permu(a) next_permutation(all(a))
#define aru(a,d) a.find(d)!=a.end()
#define nai(a,d) a.find(d)==a.end()
//#define aru p.find(mp(x,y))!=p.end()
//#define grid_input(a,type) int h,w;std::cin >> h>>w;vvector(a,type,h,w,0);reprep(i,j,h,w)std::cin >> a[i][j];

//typedef long long T;
ll ceili(ll a,ll b){
    return ((a+b-1)/b);
}
const int INF = 2000000000;
//const ll INF64 =922330720854775807ll;
//const ll INF64 = 4223372036854775807ll;
//const ll INF64 = 9223372036854775807ll;

//const ll INF64 = 243'000'000'000'000'000'0;
const ll INF64 = 430000000000000000;


const ll MOD = 1000000007ll;
//const ll MOD = 998244353ll;
//const ll MOD = 1000003ll;
const ll OD = 1000000000000007ll;
const dou pi=3.141592653589793;

//mod=MOD;

long long modpow(long long a, long long n) { 
    long long res = 1;
    a%=MOD;
    while (n > 0) {
        if (n & 1) res = res * a % MOD;
        a = a * a % MOD;
        n >>= 1;
    }
    return res;
}
vector< ll > divisor(ll n) {  //約数の列挙
  vector< ll > ret;
  for(ll i = 1; i * i <= n; i++) {
    if(n % i == 0) {
      ret.push_back(i);
      if(i * i != n) ret.push_back(n / i);
    }
  }
  sort(begin(ret), end(ret));
  return (ret);
}
map< ll, ll > prime_factor(ll n) {//素因数分解
  map< ll, ll > ret;
  for(ll i = 2; i * i <= n; i++) {
    while(n % i == 0) {
      ret[i]++;
      n /= i;
    }
  }
  if(n != 1) ret[n] = 1;
  return ret;
}

bool IsPrime(ll num)//素数
{
    if (num < 2ll) return false;
    else if (num == 2ll) return true;
    else if (num % 2ll == 0ll) return false; // 偶数はあらかじめ除く

    double sqrtNum = sqrt(num);
    for (ll i = 3ll; i <= sqrtNum; i += 2ll)
    {
        if (num % i == 0ll)
        {
            // 素数ではない
            return false;
        }
    }

    // 素数である
    return true;
}

//メモ
//ゲーム(Grundy数とか)の復習をする
//個数制限付きナップサックの復習
//戻すDP
//全方位木DPとスライド最小値
//ゲーム→パリティに注目するといいことあるかも
//理由がなければLLを使え！
//理由がなければLLを使え！
//理由がなければLLを使え！
//拡張ユークリッド
//木の直径のアルゴリズム
//修論→章ごとに完成したら指導教員通して主査にいちど見てもらうこと
//↑つまり，提出期限に余裕をもって完成する必要があるし(それはそう)，二章，3章を他よりも先にさっさとまとめて結論を示すべし．


int main(){
    int n,k;
    std::cin >> n>>k;
    string s;
    std::cin >> s;
    /*
    std::cin >> s;
  
    n=26;
    s="abcdefghijklmnopqrstuvwxyz";
    rep(i,n-k+1){
        reverse(s.begin()+i,s.begin()+i+k);
        std::cout << s << std::endl;
    }
    */
    
    
    ///*
    if(k%2==1){
        std::cout << s.substr(k-1,n-k+2) <<s.substr(0,k-1)<< std::endl;
    }
    else{
        string t=s.substr(0,k-1);
        string tt=s.substr(k-1,n-k+1);
     //   ppri(t,tt);
       reverse(all(t));
       // reverse(all(tt));
        
        std::cout << tt<<t << std::endl;
    }
    //*/
}