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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <unordered_set>
#include <unordered_map>
#include <algorithm>

using namespace std;

unsigned int list[] = {3, 5, 7};

int A(unsigned int base, unsigned int N)
{
  int count = 0;
  if(N < base) return count;
  
  if(base <= N) count++;
  unsigned int a = base*10;
  for(auto n : list){
    count += A(a + n, N);
  }
  return count;
}

int B(unsigned int base, unsigned int N, int digit)
{
  int count = 0;
  if(N < base) return count;
  
  if(base <= N) count++;
  unsigned int a = base*10;
  for(auto n : list){
    count += B(base + n * 10 * (digit-1), N, digit+1);
  }
  return count;
}

int C(unsigned int base, unsigned int N)
{
    return A(base, N) + B(base, N, 3);
}

int main()
{
  unsigned int N;
  cin >> N;
  
  int count = 0;
  count += C(357, N);
  count += C(375, N);
  count += C(537, N);
  count += C(573, N);
  count += C(735, N);
  count += C(753, N);
  
  cout << count << endl;
  return 0;
}