C. Hexadecimal's Numbers

time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

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

output

Note

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

由1 0 组成的数字，就是二进制数字的值！

把当前数字用10表示的最大数字，所代表的二进制的值就是答案。

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<sstream>

#include<algorithm>

#include<queue>

#include<deque>

#include<iomanip>

#include<vector>

#include<cmath>

#include<map>

#include<stack>

#include<set>

#include<fstream>

#include<memory>

#include<list>

#include<string>

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

#define MAXN 503

#define N 33

#define MOD 10000007

#define INF 1000000009

const double eps = 1e-;

const double PI = acos(-1.0);

string str;

int main()

{

    cin >> str;

    bool f = false;

    for (int i = ; i < str.size(); i++)

    {

        if (f) str[i] = '';

        else if (str[i] > '')

        {

            str[i] = '';

            f = true;

        }

    }

    LL ans = str[str.size() - ] - '', bas = ;

    for (int i = str.size() - ; i >= ; i--)

    {

        bas *= ;

        ans = ans + (str[i] - '')*bas;

    }

    cout << ans << endl;

    return ;

}

巴特西

C. Hexadecimal's Numbers

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