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

from sys import stdin,stdout,setrecursionlimit,maxint,exit
#setrecursionlimit(2*10**5)
def listInput():
 return map(long,stdin.readline().split())
def printBS(li):
 for i in xrange(len(li)-1):
  stdout.write("%d "%li[i])
 stdout.write("%d\n"%li[-1])
quasi=[]
#since quasi numbers can be 2^6+1
#find by bruting
def brute(s,i,n):
 if i==n+1:
  if int(s)!=0:
   #dont need 0
   quasi.append(int(s))
  return
 brute(s+"0",i+1,n)
 brute(s+"1",i+1,n)
brute("",1,6)
quasi.append(1000000)
def sin():
 return stdin.readline().rstrip()
n=input()
#solution based on dp
#a number cane be represented minimaly by subbracting 
#a quasi no from it and then solving for that
#ans is min no of quasis used in all ways
#O(n*33)
# dp equation 
dp=[(maxint,1)]*(n+1)
dp[0]=(0,0)
dp[1]=(1,1)
for i in xrange(2,n+1):
 j=0
 while j<len(quasi) and quasi[j]<=i:
  if dp[i][0]>dp[i-quasi[j]][0]+1:
   dp[i]=(dp[i-quasi[j]][0]+1,quasi[j])
  j+=1
print dp[n][0]
while n:
 print dp[n][1],
 n=n-dp[n][1]
  