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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include <stdio.h>

#define bn 10000

int is753num(int n)
{
    int line[12];
    for (int i = 0; i < 12; i++)
    {
        line[i] = n != 0 ? n % 10 : bn;
        n /= 10;
    }
    int c7, c5, c3;
    c3 = c5 = c7 = 0;
    for (int i = 0; i < 12; i++)
    {
        // printf("%d\n",line[i]);
        switch (line[i])
        {
        case 3:
            c3++;
            break;
        case 5:
            c5++;
            break;
        case 7:
            c7++;
            break;
        case bn:
            goto outofloop;
        default:
            return 0; //------------------------>>
        }
    }
outofloop:
    if (c3 > 0 && c5 > 0 && c7 > 0)
    {
        return 1;
    }
    else
    {
        return 0;
    }
}

int main(int argc, char const *argv[])
{
    int n;
    scanf("%d", &n);
    int ans=0;
    for(int i=1; i<=n; i++){
        if(is753num(i)){
            ans++;
        }
    }
    printf("%d\n", ans);
    return 0;
}