Problem: There is a puzzle to complete by combining 14 numbers from 1 to 9. Complete by adding another number to the given 13 numbers.

The conditions for completing the puzzle are

* You must have one combination of the same numbers.
* The remaining 12 numbers are 4 combinations of 3 numbers.
The combination of three numbers is either three of the same numbers or three consecutive numbers. However, sequences such as 9 1 2 are not considered consecutive numbers.
* The same number can be used up to 4 times.



Create a program that reads a string of 13 numbers and outputs all the numbers that can complete the puzzle in ascending order. If you cannot complete the puzzle by adding any number from 1 to 9, output 0.

For example, if the given string is 3456666777999

If there is a "2", 234 567 666 77 999
If there is a "3", then 33 456 666 777 999
If there is a "5", then 345 567 666 77 999
If there is an "8", then 345 666 678 77 999

And so on, the puzzle is complete when one of the numbers 2 3 5 8 is added. Note that "6" is fine, but it will be used for the 5th time, so it cannot be used in this example.



Input

The input consists of multiple datasets. For each dataset, 13 numbers are given on one line. The number of datasets does not exceed 50.

Output

For each dataset, the numbers that can complete the puzzle are output on one line in ascending order, separated by blanks.

Example

Input

3649596966777
6358665788577
9118992346175
9643871425498
7755542764533
1133557799246


Output

2 3 5 8
3 4
1 2 3 4 5 6 7 8 9
7 8 9
1 2 3 4 6 7 8
0

Solution (in CPP):

Solution:
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#define rep(i,n) for(int i=0; i<int(n); i++)
#define all(c) (c).begin(), (c).end()
using namespace std;

bool canMakeRec(vector<int> ns) {
  if (ns.size() == 0) return true;
  int first = ns[0];  // どの順番に役を作っても結果は変わらないので、まず最初の数に着目する。配列はソートされているのでこれは使える数字の中で最小になる
  if (count(all(ns), first) >= 3) {  // 同じ数字が3個以上ある
    vector<int> nsNew = ns;  // 現在の数字をコピー
    rep(i,3) nsNew.erase( find(all(nsNew), first) );  // 3個消す
    if (canMakeRec(nsNew)) return true;
  }
  if (count(all(ns), first+1) > 0 && count(all(ns), first+2)) {  // 3つの連続する数字がある
    vector<int> nsNew = ns;  // 現在の数字をコピー
    rep(i,3) nsNew.erase( find(all(nsNew), first + i) );  // 3個消す
    if (canMakeRec(nsNew)) return true;
  }
  return false;
}

bool canMake(vector<int> ns) {
  sort(all(ns));
  // まず2つ組を作る
  int prev = -1;
  rep(i, ns.size()-1) {
    if (prev != ns[i] && ns[i] == ns[i+1]) {
      vector<int> nsNew = ns;  // 現在の数字をコピー
      nsNew.erase(nsNew.begin() + i);  // 2つ消す
      nsNew.erase(nsNew.begin() + i);
      if (canMakeRec(nsNew)) return true;
    }
    prev = ns[i];
  }
  return false;
}

int main() {
  string s;
  while (cin >> s) {
    vector<int> ns(14);
    vector<int> freq(10);  // freq[i]: 数iが現れた回数
    rep(i,13) {
      ns[i] = s[i] - '0';
      freq[ns[i]]++;
    }
    vector<int> ans;

    for (int i = 1; i <= 9; i++) {
      ns[13] = i;
      if (freq[i] < 4 && canMake(ns)) ans.push_back(i);
    }

    if (ans.size() == 0) cout << 0;
    else {
      cout << ans[0];
      rep(i, ans.size()-1) cout << " " << ans[i+1];
    }
    cout << endl;
  }
  return 0;
}