#include #pragma GCC target("avx2") #pragma GCC optimize("Ofast") #pragma GCC optimize("unroll-loops") //#include using namespace std; //using namespace boost::multiprecision; //#include //#include //#include //#include //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 inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; } template 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 P; typedef pair 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<= (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 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< > name(m,vector (n,a)) //#define vvector(name,typ,m,n)vector > name(m,vector (n)) #define vvvector(name,t,l,m,n,a) vector > > name(l, vector >(m, vector(n,a))); #define vvvvector(name,t,k,l,m,n,a) vector > > > name(k,vector > >(l, vector >(m, vector(n,a)) )); //#define case std::cout <<"Case #" <> 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) <