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>
#include<string.h>
#define q(n) pow(3,n)-3*pow(2,n)+3
int main(void){
  int i,ans=0,minus=0;
  char n[10];
  scanf("%s",n);
  for(i=0;i<strlen(n)-2;i++)ans+=q(strlen(n)-i);
  if(n[0]<'8'){
    if(n[0]=='7'){
    }
    if(n[0]=='6'||n[0]=='5'){
    }
    if(n[0]=='4'||n[0]=='3'){
      if(n[1]=='7'||n[1]=='8'||n[1]=='9'){
        minus=2*q(strlen(n)-1);
      }
      if(n[1]=='6'||n[1]=='5'){
        minus=q(strlen(n)-1);
      } 
      if(n[1]=='4'||n[1]=='3'){
      }
    }
    if(n[0]=='2'||n[0]=='1')minus=q(strlen(n));
  }
    
  ans-=minus;
  printf("%d\n",ans);
  return 0;
}