Problem: IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting.

Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from 1 to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing?

A string a is lexicographically smaller than a string b if and only if one of the following holds:

  * a is a prefix of b, but a ≠ b;
  * in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.

Input

The first and the only line contains a string s (1 ≤ |s| ≤ 1000), describing the initial string formed by magnets. The string s consists only of characters 'a' and 'b'.

Output

Output exactly |s| integers. If IA should reverse the i-th prefix (that is, the substring from 1 to i), the i-th integer should be equal to 1, and it should be equal to 0 otherwise.

If there are multiple possible sequences leading to the optimal answer, print any of them.

Examples

Input

bbab


Output

0 1 1 0


Input

aaaaa


Output

1 0 0 0 1

Note

In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string.

In the second example, she can reverse any subset of prefixes — all letters are 'a'.

Solution (in JAVA):

Solution:

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;



 
public class Try
{
    
    static class InputReader {
 
		private final InputStream stream;
		private final byte[] buf = new byte[8192];
		private int curChar, snumChars;
 
		public InputReader(InputStream st) {
			this.stream = st;
		}
 
		public int read() {
			if (snumChars == -1)
				throw new InputMismatchException();
			if (curChar >= snumChars) {
				curChar = 0;
				try {
					snumChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (snumChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}
 
		public int ni() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
 
		public long nl() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
 
		public int[] nia(int n) {
			int a[] = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = ni();
			}
			return a;
		}
 
		public String rs() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}
 
		public String nextLine() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}
 
		public boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
 
		private boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}
 
	}
	

	
	
	
    static long mod=1000000007;
    static BigInteger bigInteger = new BigInteger("1000000007");
    static int n = (int)1e6;
    static boolean[] prime;
    static ArrayList<Integer> as;
    static HashSet<Long> hs;
    static void sieve() {
			Arrays.fill(prime	, true);
			prime[0] = prime[1] = false;
			for(int i = 2 ; i * i <= n ; ++i) {
				if(prime[i]) {
					for(int k = i * i; k< n ; k+=i) {
						prime[k] = false;
					}
				}
			}
			
		}
		static PrintWriter w = new PrintWriter(System.out);
 static char [][]sol;
 static int t2 = 0;
         static int t3 = 0;
    public static void main(String[] args)
    {
        
        InputReader sc = new InputReader(System.in);
        //PrintWriter w = new PrintWriter(System.out);
    
        
        prime  = new boolean[n + 1];
		sieve();
		prime[1] = false;
		
		/*
		as = new ArrayList<>();
		for(int i=2;i<=1000000;i++)
		{
		    if(prime[i])
		    as.add(i);
		}
		*/
		
       /* 
		
	long a = sc.nl();
	 
            BigInteger ans = new BigInteger("1");
            for (long i = 1; i < Math.sqrt(a); i++) {
                if (a % i == 0) {
                    if (a / i == i) {
                        ans = ans.multiply(BigInteger.valueOf(phi(i)));
                        
                    } else {
                        ans = ans.multiply(BigInteger.valueOf(phi(i)));
                        ans = ans.multiply(BigInteger.valueOf(phi(a / i)));
 
                        
                    }
 
 
                }
            }
            w.println(ans.mod(bigInteger));
            */
          
         // MergeSort ob = new MergeSort();
       // ob.sort(arr, 0, arr.length-1);
       
    // Student []st = new Student[x];    
   // st[i] = new Student(i,d[i]);
     //Arrays.sort(st,(p,q)->p.diff-q.diff);
    
 
   String s = sc.rs();
   int f = 0;
   int max = 0;
   int r = 0;
   int []res = new int[s.length()];
   int temp = 0;
   int flag = 0;
   for(int i=1;i<s.length();i++)
   {
      if(s.charAt(i) == 'a' && s.charAt(i-1) == 'b' && temp == 0)
      {
          temp = 1;
          res[i-1] = 1;
          if(i==s.length()-1)
              flag = 1;
      }
          
      if(s.charAt(i) == 'b' && s.charAt(i-1) == 'a' && temp == 1)
      {
          temp = 0;
          res[i-1] = 1;
          
      }
   }
  if(s.charAt(s.length()-1) == 'a')
      res[s.length()-1] = 1;
  // res[r] = 1;
   for(int i=0;i<s.length();i++)
   {
       w.print(res[i] + " ");
   }
    


   
     
     
   w.close();
        
  
    
    }
 static int digit(long x)
 {
     int p = 0;
     while(x > 0)
     {
         x /= 10;
         p++;
     }
     return p;
 }
 
    public static String f(String a,int b){
        if(b==0)
           return "";
       else if(b==1)
            return a;
        else{
            if(b%2==0)
               return f(a,b/2)+f(a,b/2);
            else
                return a+f(a,b/2)+f(a,b/2);
        }
            
    }
     
   
     static class Student
    {
        
        int first;
        int sec;
        Student(int first,int sec)
        {
            this.first = first;
            this.sec = sec;
            
            
        }
        
    }
     
     public static long modMultiply(long one, long two) {
 
		return (one % mod * two % mod) % mod;
	}
    static long fx(int m)
    {
        long re = 0;
        for(int i=1;i<=m;i++)
        {
            re += (long) (i / gcd(i,m));
        }
        return re;
    }
     static long gcd(long a, long b)
    {
	 if (a == 0)
	  return b;
				         
	return gcd(b%a, a);
    }
    
   
    
     static long phi(long nx)
    {
        // Initialize result as n
        double result = nx; 
        
     
        // Consider all prime factors of n and for 
        // every prime factor p, multiply result
        // with (1 - 1/p)
        for (int p = 0; as.get(p) * as.get(p) <= nx; ++p)
        {
            // Check if p is a prime factor.
            if (nx % as.get(p) == 0)
            {
                // If yes, then update n and result
                while (nx % as.get(p) == 0)
                    nx /= as.get(p);
                result *= (1.0 - (1.0 / (double) as.get(p)));
            }
        }
     
        // If n has a prime factor greater than sqrt(n)
        // (There can be at-most one such prime factor)
        if (nx > 1)
            result *= (1.0 - (1.0 / (double) nx));
     
        return (long)result;
    
        
        //return(phi((long)result,k-1));
        
    }
    
    public static void primeFactors(long n)
    {
        //hs = new HashSet<>();
       
        // Print the number of 2s that divide n
        long temp = n;
       //System.out.print(2 + " ");
            //n /= 2;
            
    
 
        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (long i = 2; i *i<= n; i++)
        {
            // While i divides n, print i and divide n
            if (n%i == 0)
            {
            
               // System.out.print(i + " ");
                
                hs.add(i);
                //n /= i;
             
                
            }
            
        }
 
        // This condition is to handle the case whien
        // n is a prime number greater than 2
        /*
        if (n >= 2 && n*n <= temp)
        {
          
            hs.add(n);
        }
        */
           
    }
    static int digitsum(int x)
    
    {    
 
        int sum = 0;
        while(x > 0)
        {
            int temp = x % 10;
            sum += temp;
            x /= 10;
        }
        return sum;
        
    }
    static int countDivisors(int n)
    {
        int cnt = 0;
        for (int i = 1; i*i <=n; i++)
        {
             if (n % i == 0 && i<=1000000)
             {
                // If divisors are equal,
                // count only one
                if (n / i == i)
                   cnt++;
      
                else // Otherwise count both
                    cnt = cnt + 2;
             }
         }
        return cnt;
    }
   
    static boolean isprime(long n)
    {
    
    if(n == 2)
        return true;
    if(n == 3)
        return true;
    if(n % 2 == 0)
        return false;
    if(n % 3 == 0)
        return false;
 
    long i = 5;
   long w = 2;
 
    while(i * i <= n)
    {
        if(n % i == 0)
            return false;
    
 
        i += w;
        w = 6 - w;
    }
 
    return true;
    }
    
    static long log2(long value) {
    return Long.SIZE-Long.numberOfLeadingZeros(value);
}
 
 
 
 
    
    
    static boolean binarysearch(int []arr,int p,int n)
    {
        //ArrayList<Integer> as = new ArrayList<>();
        //as.addAll(0,at);
        //Collections.sort(as);
        boolean re = false;
        int st = 0;
        int end = n-1;
        
        
        
        while(st <= end)
        { 
            int mid = st + (end-st)/2;
        
            if(p > arr[mid])
            {
                st = mid+1;
                
            }
            else if(p < arr[mid])
            {
                end = mid-1;
                
            }
            else if(p == arr[mid])
            {
            
            re = true;
            break;
            }
           
            
            
        }
         return re;
        
    }
    
   
    
    /* Java program for Merge Sort */
static class MergeSort
{
    // Merges two subarrays of arr[].
    // First subarray is arr[l..m]
    // Second subarray is arr[m+1..r]
    void merge(int arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        int L[] = new int [n1];
        int R[] = new int [n2];
 
        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 
    // Main function that sorts arr[l..r] using
    // merge()
    void sort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }
 
    /* A utility function to print array of size n */
    
 
   
}
    

    
    
    
    public static int ip(String s){
		return Integer.parseInt(s);
	}
	 public static String ips(int s){
		return Integer.toString(s);
	}
	
    
					 
}