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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int n;
int count(int num);
int is(int num);

int main(void){
  scanf("%d",&n);
  //ans+=count(0);
  printf("%d\n",count(0));
  return 0;
}

int count(int num){
  int ret=0;
  if(num<=n){
    if(is(num)){
      ret++;
    }
    ret+=count(num*10+3);
    ret+=count(num*10+5);
    ret+=count(num*10+7);
  }
  return ret;
}

int is(int num){
  int tmp,f1=0,f2=0,f3=0;
  while(num!=0){
    tmp=num%10;
    num=num/10;
    if(tmp==3){
      f1=1;
    }else if(tmp==5){
      f2=1;
    }else if(tmp==7){
      f3=1;
    }
  }
  if(f1*f2*f3==1)
    return 1;
  else
    return 0;
}