/* numbers.cpp - Matt Mahoney, mmahoney@cs.fit.edu

This program reads text from standard input and outputs a
list of integers extracted from the text in ascending order.
A number is any sequence of one or more digits (0-9) possibly
with a leading minus sign, but no other characters.  For example,
if the input is "-3.75-$1.05", the output is

-3
1
5
75

*/

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>   // sort()
#include <cctype>      // isdigit()
#include <cstdlib>     // atoi()
using namespace std;

int main() {
  vector<int> v;  // List of numbers extracted so far, e.g. -3, 75.
  string s;  // Characters making up a number so far, e.g. "-" or "-3"
  char c;  // Current character, e.g. '-' or '3'

  /* Extract numbers and put into v.  A number has 0 or 1 minus signs
  followed by 1 or more digits.  The following 3 x 3 state table describes
  how the input is parsed.  The 3 possible states of s are "", "-",
  and "number", representing all other possible values.  The 3 possible
  inputs are '-', a digit (0-9), or another character.

  State    Input    Next state      Action
  -----    -----    ----------      ------
  ""       '-'      "-"
  ""       digit    "digit"
  ""       other    ""
  "-"      '-'      "-"
  "-"      digit    "-digit"
  "-"      other    ""
  number   '-'      "-"             Store number in v
  number   digit    "number+digit"
  number   other    ""              Store number in v
  */

  while (cin.get(c)) {
    if (s == "") {
      if (isdigit(c) || c=='-')
        s += c;
    }
    else if (s == "-") {
      if (isdigit(c))
        s += c;
      else if (c != '-')
        s = "";
    }
    else {
      if (isdigit(c))
        s += c;
      else {
        v.push_back(atoi(s.c_str()));
        s = "";
        if (c == '-')
          s += c;
      }
    }
  }

  // Handle any leftover characters at end of file (handle as input 'other')
  if (s != "" && s != "-")
    v.push_back(atoi(s.c_str()));

  // Sort and print
  sort(v.begin(), v.end());
  for (int i=0; i<int(v.size()); ++i)
    cout << v[i] << '\n';
  return 0;
}