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: Wrong Answer)

#include<stdio.h>
#include<math.h>

int main(){
    int i, j, k;
    int N;
    int s[12];
    int n3, n5, n7;
    int num;
    int ans = 0;

    for(i=0; i<12; i++){
        s[i] = 0;
    }

    scanf("%ld", &N);

    for(i=0; i<1000000; i++){
        s[0] += 1;
        num = 0;
        n3=0; n5=0; n7=0;
        for(j=0; j<11; j++){
            if(s[j] == 4){
                s[j] = 1;
                s[j+1] += 1;
            }
            if(s[j] == 1){
                num += 3*pow(10,j);
                n3 += 1;
            }else if(s[j] == 2){
                num += 5*pow(10,j);
                n5 += 1;
            }else if(s[j] == 3){
                num += 7*pow(10,j);
                n7 += 1;
            }
        }
        if(n3*n5*n7 > 0){
            ans += 1;
        }
        if(num > N){
            break;
        }
    }

    printf("%d", ans);

    return 0;
}