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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
using namespace std;
int main(void){
    long long N,i,sum=0;
    cin >> N;
    if(N>500000000){
        sum=14384;
        for(i=500000003;i<N+1;){
        long long a=0,b=0,c=0,d=1,n=i;
        for(;n>0;){
            if(n%10==3){
                a=1;
            }else if(n%10==5){
                b=1;
            }else if(n%10==7){
                c=1;
            }else{
                d=0;
                break;
            }
            n=n/10;
        }
        if(a*b*c*d==1){
            sum++;
        }
        if(i%10==7){
            i+=6;
        }else{
            i+=2;
        }
    }
    cout << sum << endl;
    return 0;
    }
    for(i=3;i<N+1;){
        long long a=0,b=0,c=0,d=1,n=i;
        for(;n>0;){
            if(n%10==3){
                a=1;
            }else if(n%10==5){
                b=1;
            }else if(n%10==7){
                c=1;
            }else{
                d=0;
                break;
            }
            n=n/10;
        }
        if(a*b*c*d==1){
            sum++;
        }
        if(i%10==7){
            i+=6;
        }else{
            i+=2;
        }
    }
    cout << sum << endl;
}