111111
 100009
 10011
 1000000
 8
 10201
 102030
 908172
 123456
 415
 900000
 21
 909090
 314159
 999999
 909823
 987654
 1453
 1435
 98
 10
 2
 1

1
111111
 9
100001 1 1 1 1 1 1 1 1
 1
10011
 1
1000000
 8
1 1 1 1 1 1 1 1
 2
10101 100
 3
101010 1010 10
 9
101111 101011 101010 101010 101010 101010 101010 101000 100000
 6
111111 11111 1111 111 11 1
 5
111 101 101 101 1
 9
100000 100000 100000 100000 100000 100000 100000 100000 100000
 2
11 10
 9
101010 101010 101010 101010 101010 101010 101010 101010 101010
 9
111111 101011 101011 1011 11 1 1 1 1
 9
111111 111111 111111 111111 111111 111111 111111 111111 111111
 9
101111 101111 101101 101100 101100 101100 101100 101100 101000
 9
111111 111111 111111 111111 111110 111100 111000 110000 100000
 5
1111 111 111 110 10
 5
1111 111 111 101 1
 9
11 11 11 11 11 11 11 11 10
 1
10
 2
1 1
 1
1

#include <bits/stdc++.h>
using namespace std;
int N, D[1000006], F[1000006];
bool check(int n) {
  for (; n; n /= 10)
    if (n % 10 > 1) return 0;
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> N;
  vector<int> nums;
  for (int i = 1; i <= N; i++)
    if (check(i)) nums.push_back(i);
  for (int i = 1; i <= N; i++) D[i] = 2e9;
  for (int i = 0; i < N; i++)
    if (D[i] < 2e9) {
      for (int n : nums) {
        if (i + n > N) break;
        if (D[i + n] > D[i] + 1) D[i + n] = D[i] + 1, F[i + n] = i;
      }
    }
  vector<int> ans;
  for (int i = N; i; i = F[i]) ans.push_back(i - F[i]);
  printf("%d\n", ((int)(ans).size()));
  for (int t : ans) printf("%d ", t);
  puts("");
}
