. Hexadecimal's Numbers

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.

Sample test(s)

input

10

output

2

Note

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

#####################EDITORIAL#########################################

INITIALLY  PUSH 1 IN THE STACK AND NOW TRY TO ADD 10*TOP OF STACK +1 IF IT IS LESS THAN GIVEN NO  AND ALSO TRY TO PUSH 10*TOP OF STACK IN STACK IF IT IS LESS THAN GIVEN NO   NO TOTAL NO OF PUSH DONE IN THE STACK IS THE ANSWER .....

##################################CODE####################################

#include<iostream>

using namespace std;

#include<bits/stdc++.h>

int find(long long int n)

 {

    stack<long long int > sta;

      sta.push(1);

      int count =1;

         while(!sta.empty())

          {

           long long int val=sta.top();

             sta.pop();

             if(val*10+1<=n)

              {

              count++;

              sta.push(val*10+1);

               

              }

              if(val*10<=n)

               {

                  count++;

              sta.push(val*10);

               }

          }

          return count;

          

 }

int  main()

 {

  long  long int n;

   cin>>n;

    int ans=find(n);

    cout<<ans<<endl;

    return 0;

 }