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 JavaScript (node.js v5.12) (judge verdict: Accepted)

"use strict";
    
const main = arg => {
    arg = arg.trim().split("\n");
    const N = parseInt(arg[0]);
    const l = String(N).length;
    
    const num = ["3", "5", "7"];
    const X   = [3, 5, 7];
    
    let answer = 0;
    let n      = 0;
    
    // まず、3,5,7から成る数字を全て列挙する
    for(let i=0; i<l-1; i++) {
        let Xl = X.length;
        for(let j=0; j<3; j++) {
            for(let k=n; k<Xl; k++) {
                X.push(parseInt(num[j] + X[k]));
            }
        }
        
        n = Xl;
    }
    
    // 3,5,7からなる数字の配列から、それぞれが1回以上登場するものに絞る
    for(let i=0; i<X.length; i++) {
        if(X[i] > N) break;
        
        let digits = String(X[i]).split("");
        
        if(digits.indexOf("3") !== -1 && digits.indexOf("5") !== -1 && digits.indexOf("7") !== -1) {
            answer++;
        }
    }
    
    console.log(answer);
}
main(require('fs').readFileSync('/dev/stdin', 'utf8'));