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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::prelude::*;

fn input<T>() -> T
where
    T: std::str::FromStr,
{
    let stdin = std::io::stdin();
    let token = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    token.parse().ok().unwrap()
}

fn main() {
    let n = input::<i64>();

    let m = (n as f64).log10() as i64 + 1;
    let mut ans = 0;
    for i in 3..m + 1 {
        ans += dfs(0, 0, 0, 0, 0, n, i);
    }
    println!("{}", ans);
}

fn dfs(depth: i64, cur: i64, cnt7: i64, cnt5: i64, cnt3: i64, n: i64, m: i64) -> i64 {
    if depth == m {
        if cnt7 > 0 && cnt5 > 0 && cnt3 > 0 && cur <= n {
            1
        } else {
            0
        }
    } else {
        let mut cnt = 0;
        cnt += dfs(
            depth + 1,
            cur + 7 * 10i64.pow(depth as u32),
            cnt7 + 1,
            cnt5,
            cnt3,
            n,
            m,
        );
        cnt += dfs(
            depth + 1,
            cur + 5 * 10i64.pow(depth as u32),
            cnt7,
            cnt5 + 1,
            cnt3,
            n,
            m,
        );
        cnt += dfs(
            depth + 1,
            cur + 3 * 10i64.pow(depth as u32),
            cnt7,
            cnt5,
            cnt3 + 1,
            n,
            m,
        );
        cnt
    }
}