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)

/* Coached by Dukkha */
#include <stdio.h>

int count(int n) {
	return n == 0 ? 0 : count(n / 10) + 1;
}

int main() {
	int n, k, h, p3, p2, b, cnt;

	scanf("%d", &n);
	k = count(n);
	cnt = 0;
	p3 = 27, p2 = 8;
	for (h = 3; h < k; h++) {
		cnt += p3 - p2 * 3 + 3;
		p3 *= 3, p2 *= 2;
	}
	for (b = 0; b < p3; b++) {
		int a, k3, k5, k7, x;

		k3 = k5 = k7 = 0;
		x = 0;
		a = b;
		for (h = 0; h < k; h++) {
			int d = a % 3, d_;

			if (d == 0)
				k3++, d_ = 3;
			else if (d == 1)
				k5++, d_ = 5;
			else
				k7++, d_ = 7;
			x = x * 10 + d_;
			a /= 3;
		}
		if (k3 && k5 && k7 && x < n)
			cnt++;
	}
	printf("%d\n", cnt);
	return 0;
}