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>
#include <string.h>

int main(int argc, char *argv[])
{
    int  n;
	int  i,j;
	char s[101];
	int  flg3;
	int  flg5;
	int  flg7;
	int  not_n = 0;
 
	scanf("%d", &n);

    for(i=1; i<=n; i++){
        s[0] = '\0';
		sprintf(s,"%d",i);

        flg3 = 0;
		flg5 = 0;
		flg7 = 0;

		for(j=0;j<strlen(s);j++){
			if(s[j] == '7') { flg7 = 1; }
			else if (s[j] == '5') { flg5 = 1; }
			else if (s[j] == '3') { flg3 = 1; }
			else {break;}
		}

        if (flg7 == 1 && flg5 == 1 && flg3 == 1){
			not_n++;
        }
	}

    printf("%d\n",not_n);
	
	return 0;
}