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: Wrong Answer)

#include <stdio.h>

int main(void)
{
	long n, ans=0, a[1000000], k=10, k1=1;
	long b[3]={3,5,7};
	int k2=3, k3=3, c, k4=0;
	scanf("%ld", &n);
	while (k<=n) 
	{
		k*=10;
		k1++;
	}
	if (k1>3)
	{
		for (int i = 0; i < 3; ++i)
		{
			a[i]=b[i];
		}
		for (int i = 1; i < k1; ++i)
		{
			for (int i = 0; i < k2; ++i)
			{
				for (int j = 0; j < 3; ++j)
				{
					a[k3]=a[k4+i]*10+b[j];
					k3++;
				}
			}
			k4+=k2;
			k2*=3;
		}

		for (int i = 0; i < k3; ++i)
		{
			if (a[i]>n) break;
			c=1;
			while (a[i] > 0)
			{
				c=c*((a[i]+10)%10);
				a[i]/=10;
			}
			if (c%105==0)
			{
				ans+=1;
			}
		}
	}
	printf("%ld\n", ans);
}