codeforces 9C 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.

Example

Input

10

Output

2

Note

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

思路：既然是0 1 组成的整数，那么直接可以枚举所有由01组成的整数就可以了，但是不能盲目枚举，从小到达枚举即可，就算是最大的数最多只有9位，枚举九位的个数为2^9-1，所以最多也不过513个数，暴力枚举稳过的。

代码如下：

#include <cstdio>int sum;void dfs(int x , int y) {if(x <= y)sum++;else return ;dfs(x * 10 , y);dfs(x * 10 + 1 , y);}int main() {int n;scanf("%d" , &n);sum = 0;dfs(1 , n);printf("%d\n" , sum);return 0;}