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>
#define int long long

int O357[4] = {0, 3, 5, 7};

int convert(int n){
	int ans, i;
	for(ans = 0, i = 1; n > 0; n /= 4, i *= 10){
		ans += i * O357[n % 4];
	}
	return ans;
}

int is_753(int n){
	int f3, f5, f7;
	for(f3 = 0, f5 = 0, f7 = 0; n > 0; n /= 10){
		if(n % 10 == 3){
			f3 = 1;
		}
		else if(n % 10 == 5){
			f5 = 1;
		}
		else if(n % 10 == 7){
			f7 = 1;
		}
		else{
			return 0;
		}
	}
	if(f3 == 1 && f5 == 1 && f7 == 1){
		return 1;
	}
	else{
		return 0;
	}
}

signed main(){
	int N, i, ans = 0;
	scanf("%lld", &N);
	for(i = 1; convert(i) <= N; i++){
		if(is_753(convert(i)) == 1){
//			printf("convert(%d) = %d\n", i, convert(i));
			ans++;
		}
	}
	printf("%lld\n", ans);
	return 0;
}