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: Accepted)

#include <stdio.h>

long long N;

int dfs(long long i){
  //printf("i=%lld\n",i);
  if(i>N) return 0;
  //自分が条件を満たすか調べる
  int count;
  int is_ok; //自身が満たすなら1,満たさないなら0.
  int count3=0,count5=0,count7=0;
  long long j;
  j=i;
  while(j!=0){
    if(j%10==3) count3++;
    else if(j%10==5) count5++;
    else if(j%10==7) count7++;
    j=j/10;
  }

  if(count3>0 && count5>0 && count7>0) is_ok=1;
  else is_ok=0;

  count=dfs(10*i+3)+dfs(10*i+5)+dfs(10*i+7)+is_ok;
  return count;
}

int main(){
  long long NN;
  scanf("%lld",&NN);
  N=NN;
  int count;
  count=dfs(0);

  printf("%d\n",count);
}