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 <bits/stdc++.h>
using namespace std;

int char_count(string s, char c) {
  int cnt;
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == c) cnt++;
  }
  return cnt;
}

int main() {
  unsigned long n, count = 1, th, fi, se;
  string num;
  cin >> n;
  if (n < 357) {
    cout << "0" << endl;
    return 0;
  }
  else if (n == 357) {
    cout << count << endl;
    return 0;
  }
  for (int i = 358; i <= n; i++) {
    num = to_string(i);
    th = char_count(num, '3');
    fi = char_count(num, '5');
    se = char_count(num, '7');
    if (th>0 && fi>0 && se>0 && (th+fi+se) == num.length()) count += 1;
  }
  cout << count << endl;
  return 0;
}