#include <algorithm>
#include <iostream>
#include <vector>

using namespace std;
using LL = long long;

const int kN = 5e5 + 1;

struct SAM {
  struct E {
    int n[26], ml, pr, c, v;
    LL d;
    vector<int> e;
  } e[kN << 1];
  int c, l;

  SAM() { e[0].pr = -1; }
  void I(char ch) {
    ch -= 'a';
    int x = ++c, p = l;
    e[x].ml = e[l].ml + 1, e[x].c = 1;
    for (; ~p && !e[p].n[ch]; e[p].n[ch] = x, p = e[p].pr) {
    }
    if (~p) {
      int q = e[p].n[ch];
      if (e[q].ml == e[p].ml + 1) {
        e[x].pr = q;
      } else {
        int _q = ++c;
        copy(e[q].n, e[q].n + 26, e[_q].n);
        e[_q].ml = e[p].ml + 1, e[_q].pr = e[q].pr, e[q].pr = _q;
        for (; ~p && e[p].n[ch] == q; e[p].n[ch] = _q, p = e[p].pr) {
        }
        e[x].pr = _q;
      }
    }
    l = x;
  }
  int D(int x) {
    for (int i : e[x].e) {
      e[x].c += D(i);
    }
    return e[x].c;
  }
  void Sc() {
    for (int i = 1; i <= c; ++i) {
      e[e[i].pr].e.push_back(i);
    }
    D(0);
  }
  LL Sd(int x, int ty) {
    if (e[x].d) {
      return e[x].d;
    }
    if (x) {
      e[x].d = e[x].v = (ty ? e[x].c : 1);
    }
    for (int i = 0; i < 26; ++i) {
      int y;
      if (y = e[x].n[i]) {
        e[x].d += Sd(y, ty);
      }
    }
    return e[x].d;
  }
  string Kth(LL k, int ty) {
    if (k >= e[0].d) {
      return "-1";
    }
    int x = 0;
    string s = "";
    for (; k;) {
      for (int i = 0; i < 26; ++i) {
        int y;
        if (y = e[x].n[i]) {
          LL v = e[y].d;
          if (v >= k) {
            s += i + 'a', x = y, k -= e[y].v;
            break;
          } else {
            k -= v;
          }
        }
      }
    }
    return s;
  }
} t;
string s;
int ty, k;

int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> s >> ty >> k;
  for (char c : s) {
    t.I(c);
  }
  t.Sc(), t.Sd(0, ty);
  cout << t.Kth(k, ty);
  return 0;
}

我的这份代码可以通过此题，但是可以被下面这组数据 hack，原因是当 k==e[0].d 时是存在第 $k$ 小子串的，但我的程序误以为不存在。

input:
aaa
0 3

answer:
aaa

my_output:
-1