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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
int pow (int a,int b)
{
	int i;
	int z = 1;
	for (i = 1;i <= b;i++)
	{
		z = z * a;
	}
	return (z);
}
int main(void)
{
	long long int Number;
	scanf("%lld",&Number);
	int j;
	int Digit[10] = {0};
	int Final = 0;
	for (j = 1;j <= Number;j++)
	{
		int i;
		Digit[0] = j % 10;
		int digit = 0;
		for (i = 1;i <= 10;i++)
		{
			if (j / pow (10,i) == 0)
			{
				digit = i;
				break;
			}
		}
		for (i = 1;i < 10;i++)
		{
			Digit[i] = ((j % pow (10,i + 1)) - Digit[i - 1]) / pow (10,i);
		}
		int Num3 = 0;
		int Num5 = 0;
		int Num7 = 0;
		int Num = 0;
		for (i = 0;i < digit;i++)
		{
			if (Digit[i] == 3)
			{
				Num3++;
			}
			else if (Digit[i] == 5)
			{
				Num5++;
			}
			else if (Digit[i] == 7)
			{
				Num7++;
			}
			else
			{
				Num++;
			}
		}
		if (Num3 && Num5 && Num7 && (Num == 0))
		{
			Final++;
		}
		else
		{
			continue;
		}
	}
	printf("%d\n",Final);
}