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<math.h>

int is753(long long int number) {
  long long int digit = 1;
  int f3 , f5, f7;
  f3 = f5 = f7 = 0;
  long long int tmp = number;
  while(tmp > 0) {
    int n = tmp % 10;
    if(n == 3) {
      f3 = 1;
    } else if(n==5) {
      f5 = 1;
    } else if(n==7) {
      f7 = 1;
    } else {
      return digit;
    }
    digit *= 10;
    tmp /= 10;
  }
  return (f3 && f5 && f7) ? 0 : 1;
}

int main(void) {
  long long int N;
  scanf("%lld", &N);

  int count = 0;
  long long int i = 0;
  while( i < N ) {
    int r = is753(i);
    if(r == 0) {
      count ++;
      i++;
    } else {
      i += r;
    }
  }

  printf("%d\n", count);

  return 0;
}