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: Runtime Error)

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

int main(){

  long long int d=1000, n;
  int i, j, k, ans=0, check[10]={};
  char num[20];
  char e[20]="777";
  scanf("%lld", &n);
  for(i=0; i<=9; i++){
    check[i]=0;
  }
  for(i=3; i<=9; i++){
    for(j=0; j<i; j++){
      num[j] = '3';
    }
    while(strcmp(num,e)!=0){
      for(j=0;j<i;j++){
        check[num[j]-'0']++;
      }
      if(check[3]>0 && check[5]>0 && check[7]){
        if(atoi(num)<=n){
          ans++;
        }
      }
      check[3]=0;
      check[5]=0;
      check[7]=0;
      k=0;
      for(j=0; k==0; j++){
        if(num[j]=='7'){
          num[j]='3';
          k=0;
        }
        else if(num[j]=='5'){
          num[j]='7';
          k=1;
        }
        else if(num[j]=='3'){
          num[j]='5';
          k=1;
        }
      }
    }
    e[i]='7';
    d = d*10;
  }
  printf("%d\n", ans);
  return 0;
}