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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;

class Program
{
    static void Main(string[] args)
    {
        var N = long.Parse(Console.ReadLine());
        var count = dfs("0", N);
        Console.WriteLine(count);
    }

    static long dfs(string current, long N)
    {
        var cur = long.Parse(current);
        if (cur > N)
        {
            return 0;
        }

        bool hasSeven = false;
        bool hasFive = false;
        bool hasThee = false;
        foreach (var ch in current)
        {
            if (ch == '7')
            {
                hasSeven = true;
            }
            else if (ch == '5')
            {
                hasFive = true;
            }
            else if (ch == '3')
            {
                hasThee = true;
            }
        }

        long count = 0;

        if (hasSeven && hasFive && hasThee)
        {
            count++;
        }

        var addValues = new string[] { "7", "5", "3" };
        foreach (var addValue in addValues)
        {
            count += dfs(current + addValue, N);
        }
        return count;
    }
}