One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.

But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.

Input

Input data contains the only number n (1 ≤ n ≤ 109).

Output

Output the only number — answer to the problem.

Examples

Input

10


Output

2

Note

For n = 10 the answer includes numbers 1 and 10.

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

public class Main { 
    
    public static void main(String[] args) throws java.lang.Exception { 
        Reader pm =new Reader(); 
        //Scanner pm = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        long n = pm.nextLong();
        ArrayList<String> al = generatePrintBinary(1024);
        //System.out.println(al);
        int ans = 0;
        
        for(int i=0;i<al.size();i++){
            long pl = Long.parseLong(al.get(i));
            if(n>=pl)
                ans++;
            else
                break;
        }
        System.out.println(ans);
    }
    
    static ArrayList<String> generatePrintBinary(int n) 
	{ 
	    ArrayList<String> al = new ArrayList<>();
		// Create an empty queue of strings 
		Queue<String> q = new LinkedList<String>(); 
		
		// Enqueue the first binary number 
		q.add("1"); 
		
		// This loops is like BFS of a tree with 1 as root 
		// 0 as left child and 1 as right child and so on 
		while(n-- > 0) 
		{ 
			// print the front of queue 
			String s1 = q.peek(); 
			q.remove(); 
			//Long pl = Long.parseLong(s1,2);
			al.add(s1);
			//System.out.println(s1); 
			
			// Store s1 before changing it 
			String s2 = s1; 
			
			// Append "0" to s1 and enqueue it 
			q.add(s1 + "0"); 
			
			// Append "1" to s2 and enqueue it. Note that s2 contains 
			// the previous front 
			q.add(s2 + "1"); 
		} 
		return al;
	} 
	
    static class Reader { 
        final private int BUFFER_SIZE = 1 << 16; 
        private DataInputStream din; 
        private byte[] buffer; 
        private int bufferPointer, bytesRead; 
  
        public Reader() 
        { 
            din = new DataInputStream(System.in); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public Reader(String file_name) throws IOException 
        { 
            din = new DataInputStream(new FileInputStream(file_name)); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public String readLine() throws IOException 
        { 
            byte[] buf = new byte[64]; // line length 
            int cnt = 0, c; 
            while ((c = read()) != -1) 
            { 
                if (c == '\n') 
                    break; 
                buf[cnt++] = (byte) c; 
            } 
            return new String(buf, 0, cnt); 
        } 
  
        public int nextInt() throws IOException 
        { 
            int ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do
            { 
                ret = ret * 10 + c - '0'; 
            }  while ((c = read()) >= '0' && c <= '9'); 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public long nextLong() throws IOException 
        { 
            long ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public double nextDouble() throws IOException 
        { 
            double ret = 0, div = 1; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
  
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
  
            if (c == '.') 
            { 
                while ((c = read()) >= '0' && c <= '9') 
                { 
                    ret += (c - '0') / (div *= 10); 
                } 
            } 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        private void fillBuffer() throws IOException 
        { 
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
            if (bytesRead == -1) 
                buffer[0] = -1; 
        } 
  
        private byte read() throws IOException 
        { 
            if (bufferPointer == bytesRead) 
                fillBuffer(); 
            return buffer[bufferPointer++]; 
        } 
  
        public void close() throws IOException 
        { 
            if (din == null) 
                return; 
            din.close(); 
        } 
    }
}    