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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void){
  int j,l,other;
  long N,i,count7,count5,count3,count;
  char str[100],s[100];
  
  fgets(str,sizeof(str),stdin);
  N=atol(str);
  
  for(i=1,count=0;i<N+1;i++){
    sprintf(s,"%ld",i);
    l=strlen(s);
    for(j=0,count7=0,count5=0,count3=0,other=0;j<l;j++){
      if(s[j]=='7'){
        count7++;
      }else if(s[j]=='5'){
        count5++;
      }else if(s[j]=='3'){
        count3++;
      }else{
        other++;
        break;
      }
    }
    if(count7>0 && count5>0 && count3>0 && other==0){
      count++;
    }
  }
  
  printf("%ld\n",count);
  
  return 0;
}