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: Time Limit Exceeded)

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <string>
#include <vector>
#include <set>
#include <stack>
using namespace std;

bool divide(int n)
{
    int cnt_other = 0, cnt3 = 0, cnt5 = 0, cnt7 = 0;
    while(n > 0){
        int tmp = n % 10;
        switch(tmp) {
			case 3: cnt3++; break;
			case 5: cnt5++; break;
			case 7: cnt7++; break;
			default: cnt_other++; break;
		}
        n /= 10;
    }
    if (cnt_other == 0 && cnt3 > 0 && cnt5 > 0 && cnt7 > 0) {
        return true;
    } else {
        return false;
    }
}

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);
    int n; cin >> n;

    int cnt = 0;
    for (int i = 1; i <= n; i++) {
        if (divide(i)) {
            cnt++;
        }
    }
    printf("%d\n", cnt);
	return 0;
}