Finding lexicographically k'th smallest substring

3. Finding lexicographically k'th smallest substring:
                                                     The number of path / substring
  of a node / state of suffix automaton defines that we can obtain such number of
  substring which starts with the transitions. So, at first we calculate number of
  substring of all states and then do following:

  Let, there are two states which we can go from initial state. There can be two
  probabilities.
  1. The number of substring from start(1) is less than k. So, we can be sure that
     our desired substring will not start with transition(1). So, we subtract
     that number of substring from k.
  2. If the number is equal or greater then we go to these state and print the
     transition name. Then subtract 1 from k as we take 1 character. The process
     end when k is 0.

void kLex(int ver, int k)
{
    if (k == 0)
        return;
    for (auto it : st[ver].next)
    {
        int tSub = d[it.ss];
        if (tSub < k)           // this state node belong
        {
            k -= tSub;
        }
        else
        {
            cout << it.ff;
            if (k == 1)     cout << endl;

            kLex(it.ss, k-1);
            return;           // do nothing when return
        }
    }
}

Practice Problem :
                 Spoj : SUBLEX - Lexicographical Substring Search