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: Runtime Error)

#include <stdio.h>


int quasi(int *a, int *N);	//Nまでの準七五三数を列挙

int main(void){
	int N,i,j;
	scanf("%d",&N);
	
	int a[N],len;
	len=quasi(a,&N);
	//printf("%d\n",len);
	
	/*for(i=0;i<len;++i){
	printf("%d ",a[i]);
	}*/

	int three,five,seven,r,q,ans=0;
	for(i=0;i<len;++i){
		three=0,five=0,seven=0;
		q=a[i];
		for(j=0;j<len;++j){	
			r=q%10;
			q/=10;
			if(r==3) three++;
			else if(r==5) five++;
			else if(r==7) seven++;
			else break;
		}
		if(three>=1 && five>=1 && seven>=1) ans++;
	}
	printf("%d\n",ans);
	
	return 0;
}

int quasi(int *a, int *N){
	int i,j,k,n=-1;
	for(i=3;i<=7;i=i+2){
		for(j=3;j<=7;j=j+2){
			for(k=3;k<=8;k=k+2){
				n++;
				a[n]=i*100+j*10+k;
			}
		}
	}
	i=0,k=27;
	while(a[k]<*N){
		a[k]=a[i]*10+3;
		if(a[k]>=*N) break;
		k++;
		a[k]=a[i]*10+5;
		if(a[k]>=*N) break;
		k++;
		a[k]=a[i]*10+7;
		if(a[k]>=*N) break;
		k++;
		i++;
	}
	
	
	return k;
}