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: Accepted)

#include <cstdio>
#include <cmath>

long long dp[11][2][8];

int main()
{
  char s[12];
  scanf("%s", s);
  dp[0][0][0] = 1;
  int i;
  for (i = 0; s[i]; ++i)
  {
    int D = s[i] - '0';
    for (int j = 0; j < 2; ++j)
    {
      for (int m = 0; m < (1 << 3); ++m)
      {
        for (int k = 0; k <= (j ? 9 : D); ++k)
        {
          if (k != 3 && k != 5 && k != 7) continue;
          dp[i + 1][j || k < D][m | (k == 3 ? 1 : k == 5 ? 2 : 4)] += dp[i][j][m];
        }
      }
    }
  }
  long long ans = 0;
  for (long long j = 3; j < (long long)i; ++j)
  {
    ans += (long long)std::pow(3, j) - 3 * (long long)std::pow(2, j) + 3;
  }
  printf("%lld\n", dp[i][0][7] + dp[i][1][7] + ans);
  return 0;
}