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

#[allow(dead_code)]
fn with_stdout_lock<F: FnOnce(std::io::BufWriter<std::io::StdoutLock>)>(f: F) {
    let out = std::io::stdout();
    let w = std::io::BufWriter::new(out.lock());
    f(w);
}

#[allow(unused_macros)]
macro_rules! debug {
    ($var:expr) => {
        match &$var { v => eprintln!(concat!(stringify!($var), "={:?}"), v) }
    };
    ($var:expr, $($vars:expr),*) => {
        match &$var { v => eprint!(concat!(stringify!($var), "={:?} "), v) }
        debug!($($vars),*);
    };
}

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

    println!("{}", dfs(0, n, false, false, false));
}

fn dfs(x: i64, n: i64, has7: bool, has5: bool, has3: bool) -> i64 {
    if x > n {
        0
    } else {
        let mut ans = if has7 && has5 && has3 { 1 } else { 0 };
        ans += dfs(x * 10 + 7, n, true, has5, has3);
        ans += dfs(x * 10 + 5, n, has7, true, has3);
        ans += dfs(x * 10 + 3, n, has7, has5, true);
        ans
    }
}