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>

typedef long long ll;

int max(int a, int b){
  if(a > b) return a;
  else return b;
}
int min(int a, int b){
  if(a < b) return a;
  else return b;
}

int main(){

  int n;
  scanf("%d",&n);

  int count = 0;
  int tmp = 0;
  int flag,flag3,flag5,flag7;
  for(int i = 357; i <= n; ++i){
    tmp = i;
    flag = 0;
    flag3 = 0;
    flag5 = 0;
    flag7 = 0;
    while(tmp > 0){

      if(tmp % 10 != 3 && tmp % 10 != 5 && tmp % 10 != 7 ){
        flag = 1;
      }
      else if(tmp % 10 == 3) flag3 = 1;
      else if(tmp % 10 == 5) flag5 = 1;
      else if(tmp % 10 == 7) flag7 = 1;
      tmp /= 10;
    }
    if(flag == 0 && flag3 == 1 && flag5 == 1 && flag7 == 1) count ++;
  }

  printf("%d",count);
  return 0;

}