/*

     贪心水题：首先，最少的个数为n最大的一位数字mx，因为需要用1累加得到mx，

             接下来mx次循环，若是0，输出0；若是1，输出1，s[j]--；

     注意：之前的0的要忽略

 */

 #include <cstdio>

 #include <iostream>

 #include <cstring>

 #include <string>

 #include <algorithm>

 #include <cmath>

 #include <set>

 #include <map>

 using namespace std;

 const int MAXN = 1e4 + ;

 const int INF = 0x3f3f3f3f;

 int main(void)        //Codeforces Round #300 A Cutting Banner

 {

     //freopen ("B.in", "r", stdin);

     char s[];

     while (scanf ("%s", &s) == )

     {

         int len = strlen (s);

         int mx = -;

         for (int i=; i<len; ++i)    mx = max (mx, s[i] - '');

         printf ("%d\n", mx);

         for (int i=; i<=mx; ++i)

         {

             bool ok = false;

             for (int j=; j<len; ++j)

             {

                 if (s[j] == '')

                 {

                     if (!ok)    continue;

                     else    printf ("");

                 }

                 else

                 {

                     ok = true;

                     printf ("");    s[j]--;

                 }

             }

             if (i < mx)    printf (" ");

         }

         puts ("");

     }

     return ;

 }