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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace Atcoder
{
	class Program
	{
		static void Main(string[] args)
		{
			var N = int.Parse(Console.ReadLine());

			if (N < 357)
			{
				Console.Write("0");
			}
			int result = 0;
			for (int i = 357; i < N; i++)
			{
				if (i.ToString().Contains("0") || i.ToString().Contains("1") || i.ToString().Contains("2")
					|| i.ToString().Contains("4") || i.ToString().Contains("6") || i.ToString().Contains("8") || i.ToString().Contains("9"))
				{
					continue;
				}
				if (i.ToString().Contains("3") && i.ToString().Contains("5") && i.ToString().Contains("7"))
				{
					result++;
				}
			}
			Console.Write(result.ToString());
			//var N = int.Parse(firstLine[0]);
			//var K = int.Parse(firstLine[1]);

			//List<int> trees = new List<int>();
			//for (int i = 0; i < N; i++)
			//{
			//	trees.Add(int.Parse(Console.ReadLine()));
			//}
			//trees = trees.OrderBy(x => x).ToList();
			//var len = trees.Count;

			//List<int> difList = new List<int>();
			//difList.Add(0);
			//for (int i = 0; i < len-1; i++)
			//{
			//	difList.Add(trees[i + 1] - trees[i]);
			//}
			//difList = difList.OrderBy(x => x).ToList();

			//int result = 0;
			//for (int i = 0; i < K; i++)
			//{
			//	result += difList[i]; 
			//}

		}
	}
}