Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.

Input

The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).

Output

Output the decoded ternary number. It can have leading zeroes.

Examples

Input

.-.--


Output

012

Input

--.


Output

20

Input

-..-.--


Output

1012
/*
Codeforces Problem 32B - Borze
*/
import java.io.*;
import java.util.*;
import java.lang.*;



public class borze{

    static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 

    public static void main(String [] args){
        FastReader s = new FastReader();

        // 0 == . 
        // 1 == -.
        // 2 == --
        String c = s.next();
        StringBuffer ans = new StringBuffer("");

        for(int i = 0 ; i < c.length(); i++){
            if(c.charAt(i) == '.'){
                ans.append("0");
            }else if(c.charAt(i)== '-' && c.charAt(i+1) == '.'){
                ans.append("1");
                i++;
            }else if(c.charAt(i)== '-' && c.charAt(i+1) == '-'){
                ans.append("2");
                i++; //cheeky trick
            }
        }


        System.out.println(ans);
        /*
        My first ugly solution
        char[] chars = s.next().toCharArray();
        StringBuffer ans = new StringBuffer("");
        if(chars.length < 2){
            if(chars[0] == '.'){
                ans.append("0");
            }
        }
        //truly ugly code
        else{
            for(int i = 0 ; i< chars.length-1; i++){
                if(chars[i] == '.'){
                    ans.append("0");
                    if(chars[i+1] == '.'){
                        ans.append("0");
                    }
                    
                }else{
                    if(chars[i] == '-'){
                        if(chars[i+1] == '.'){
                            ans.append("1");
                            i++; //adding the cheeky i++
                        }else{
                            ans.append("2");
                            i++; //adding the cheeky i++
                        }
                    }
                }
            }
            //this one actually makes sense but this is bad brute force :(
            if((chars[chars.length - 2] == '-' && chars[chars.length-1] == '.') || (chars[chars.length - 2] == '.' && chars[chars.length-1] == '.') ){
                if(chars.length > 2)
                    ans.append("0");
            }

        }
        System.out.println(ans.toString());
        */
    }






}