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: Accepted)

#include <stdio.h>
#include <stdlib.h>

int main()
{
	long long N;
	scanf("%lld", &N);
	
	int i, count = 0, digit[10] = {7, 5, 3}, flag[3];
	long long n = 357;
	while (n <= N) {
		count++;
		while (1) {
			for (i = 0; digit[i] == 7; i++);
			if (digit[i] == 0) {
				for (; i >= 2; i--) digit[i] = 3;
				digit[1] = 5;
				digit[0] = 7;
				break;
			} else {
				for (digit[i--] += 2; i >= 0; i--) digit[i] = 3;
				for (i = 0; i < 3; i++) flag[i] = 0;
				for (i = 0; digit[i] > 0; i++) flag[(digit[i]-3)/2] = 1;
				if (flag[0] == 1 && flag[1] == 1 && flag[2] == 1) break;
			}
		}
		for (i = 9, n = 0; i >= 0; i--) n = n * 10 + digit[i];
	}
	
	printf("%d\n", count);
	fflush(stdout);
	return 0;
}