Problem:
Score : 300 points

Problem Statement

You are given an integer N. Among the integers between 1 and N (inclusive), how many Shichi-Go-San numbers (literally "Seven-Five-Three numbers") are there?

Here, a Shichi-Go-San number is a positive integer that satisfies the following condition:

When the number is written in base ten, each of the digits 7, 5 and 3 appears at least once, and the other digits never appear.

Constraints

1 \leq N < 10^9

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of the Shichi-Go-San numbers between 1 and N (inclusive).

Sample Input 1

575

Sample Output 1

4

There are four Shichi-Go-San numbers not greater than 575: 357, 375, 537 and 573.

Sample Input 2

3600

Sample Output 2

13

There are 13 Shichi-Go-San numbers not greater than 3600: the above four numbers, 735, 753, 3357, 3375, 3537, 3557, 3573, 3575 and 3577.

Sample Input 3

999999999

Sample Output 3

26484

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int check753(int s){
    int n,seven=0,five=0,three=0;
    while(seven==0||five==0||three==0){
        if(s==0) return 0;
        n=s%10;
        s/=10;
        if(n==7){
            seven=1;
        }else if(n==5){
            five=1;
        }else if(n==3){
            three=1;
        }
    }
    return 1;
}

int count753(int n,int s){
    if(s>n) return 0;
    int ret;
    if(check753(s)){
        ret=1;
    }else{
        ret=0;
    }
    if(s*10+3>0){
        ret+=count753(n,s*10+3);
    }
    if(s*10+5>0){
        ret+=count753(n,s*10+5);
    }
    if(s*10+7>0){
        ret+=count753(n,s*10+7);
    }
    return ret;
}

int main(void){
    int n;
    scanf("%d",&n);
    printf("%d\n", count753(n,0));
    return 0;
}