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<stdio.h>
//#include<algorithm>
//#include<vector>
#include<math.h>
#include<map>
using namespace std;

bool judge(long long x){
	int f3=0,f5=0,f7=0;
	int a;
	while(x!=0){
		a=x%10;
		x=x/10;
		if(a==3) f3=1;
		else if(a==5) f5=1;
		else if(a==7) f7=1;
		else return false;
		if(f3==1&&f5==1&&f7==1) return true;
	}
	//if(f3==1&f5==1&&f7==1) return true;
    return false;
}


int main(){
	map<long long,int> mp;
	long long t;
	int num;
	while(scanf("%lld",&t)==1){
		num=0;
		if(t>=357){
			for(long long i=357;i<=t;i++){
				if(i%2==1&&mp[i]!=2){
					if(judge(i)==true) num++;
					else{
						int p=0;
						while(i!=0){
							i/=10;
							p++;
						}
					for(long long j=i;j<=t;j+=1*pow(10,p)) mp[j]=2;
					}
				}
			}
		}
		printf("%d\n",num);
	}
	return 0;
}