Problem:
Score : 1300 points

Problem Statement

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.

Sample Input 1

5 1
bacba

Sample Output 1

aabca

When S=bacba, T=abcab, U=bacbaabcab, and the optimal choice of S' is S'=aabca.

Sample Input 2

10 2
bbaabbbaab

Sample Output 2

aaaabbaabb

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
constexpr long long INF = 1LL << 60;
constexpr long long MOD = 1000000007;
double PI = acos(-1.0);
#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define rep1(i, n) for (ll i = 1; i <= (n); ++i)
#define rrep(i, n) for (ll i = (n - 1); i >= 0; --i)
#define perm(c) sort(ALL(c));for(bool c##p=1;c##p;c##p=next_permutation(ALL(c)))
#define ALL(obj) (obj).begin(), (obj).end()
#define RALL(obj) (obj).rbegin(), (obj).rend()
#define pb push_back
#define to_s to_string
#define len(v) (ll)v.size()
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define print(x) cout << (x) << '\n'
#define drop(x) cout << (x) << '\n', exit(0)
#define debug(x) cout << #x << ": " << (x) << '\n'
using namespace std;
using ll = long long;
typedef pair<ll, ll> P;
typedef vector<ll> vec;
typedef vector<vector<ll>> vec2;
typedef vector<vector<vector<ll>>> vec3;
template<class S, class T> inline bool chmax(S &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class S, class T> inline bool chmin(S &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
inline ll msb(ll v) { return 1 << (31 - __builtin_clzll(v)); }
inline ll devc(ll x, ll y) { return (x + y - 1) / y; }
inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline ll lcm(ll a, ll b) { return a * (b / gcd(a, b)); }
 
struct IoSetup {
    IoSetup() {
        cin.tie(nullptr);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(10);
        cerr << fixed << setprecision(10);
    }
} iosetup;
 
template< typename T1, typename T2 >
ostream &operator << (ostream &os, const pair< T1, T2 > &p) {
    os << p.first << " " << p.second;
    return os;
}
 
template< typename T1, typename T2 >
istream &operator >> (istream &is, pair< T1, T2 > &p) {
    is >> p.first >> p.second;
    return is;
}
 
template< typename T1, typename T2, typename T3 >
ostream &operator << (ostream &os, const tuple< T1, T2, T3 > &t) {
    os << get<0>(t) << " " << get<1>(t) << " " << get<2>(t);
    return os;
}
 
template< typename T1, typename T2, typename T3 >
istream &operator >> (istream &is, tuple< T1, T2, T3 > &t) {
    is >> get<0>(t) >> get<1>(t) >> get<2>(t);
    return is;
}
 
template< typename T >
ostream &operator << (ostream &os, const vector< T > &v){
    for (int i = 0; i < (int)v.size(); ++i) {
        os << v[i] << (i + 1 != v.size() ? " " : "");
    }
    return os;
}
 
template< typename T >
istream &operator >>  (istream &is, vector< T > &v){
    for(T &in : v) is >> in;
    return is;
}

template< typename T >
ostream &operator << (ostream &os, const set< T > &st){
    int ct = 0;
    for(auto& s : st) cout << s << (++ct != st.size() ? " " : "");
    return os;
}

template <typename T>
constexpr set<T> &operator|= (set<T> &st1, const set<T> &st2) { 
    for(auto& s : st2) st1.insert(s);
    return st1; 
}

template <typename T>
constexpr set<T> &operator-= (set<T> &st1, const set<T> &st2) {
    for(auto& s : st2) if(st1.count(s)) st1.erase(s);
    return st1;
}

template <typename T>
constexpr set<T> &operator&= (set<T> &st1, const set<T> &st2) {
    auto itr = st1.begin();
    while(itr != st1.end()){
        if(!st2.count(*itr)) itr = st1.erase(itr);
        else ++itr;
    }
    return st1;
}

template <typename T>
constexpr set<T> operator| (const set<T> &st1, const set<T> &st2) {
    set<T> res = st1;
    res |= st2;
    return res;
}

template <typename T>
constexpr set<T> operator- (const set<T> &st1, const set<T> &st2) {
    set<T> res = st1;
    res -= st2;
    return res;
}

template <typename T>
constexpr set<T> operator& (const set<T> &st1, const set<T> &st2) {
    set<T> res = st1;
    res &= st2;
    return res;
}

/*---------------------------------    Tools    ------------------------------------------*/
template< typename T >
vector<T> cumsum(const vector<T> &X){
    vector<T> res(X.size() + 1, 0);
    for(int i = 0; i < X.size(); ++i) res[i + 1] += res[i] + X[i];
    return res;
}
 
template< typename S, typename T, typename F>
pair<T, T> bisearch(S left, T right, F f) {
    while(abs(right - left) > 1){
        T mid = (right + left) / 2;
        if(f(mid)) right = mid;
        else left = mid;
    }
    return {left, right};
}
 
template< typename S, typename T, typename F>
double trisearch(S left, T right, F f, int maxLoop = 90){
    double low = left, high = right;
    while(maxLoop--){
        double mid_left = high / 3 + low * 2 / 3;
        double mid_right = high * 2 / 3 + low / 3;
        if(f(mid_left) >= f(mid_right)) low = mid_left;
        else high = mid_right;
    }
    return (low + high) * 0.5;
}

template< typename F >
ll ternarySearch(ll L, ll R, F f) { //[L, R)
    ll lo = L - 1, hi = R - 1;
    while (lo + 1 != hi) {
        ll mi = (lo + hi) / 2;
        if (f(mi) <= f(mi + 1)) hi = mi;
        else lo = mi;
    }
    return hi;
}

/*------------------------------- Main Code Here -----------------------------------------*/
 
int main()
{
    ll N, K;
    cin >> N >> K;
    string S;
    cin >> S;

    string now = S;
    chmin(K, 2 * N);
    while(K--){ 
        string tmp = now;
        reverse(ALL(now));
        tmp += now;

        if(K) reverse(ALL(tmp));

        string comp(N, 'z');

        rep(i, 2 * N) if(i + N < 2 * N) {
            string x = tmp.substr(i, N);
            chmin(comp, x);
        }
        if(K) reverse(ALL(comp));
        now = comp;
    }

    print(now);


    return 0;
}