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: Wrong Answer)

#include<bits/stdc++.h>                
using namespace std;
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); i++)
#define rep1(i,n) for(int i = 0; i <= (n); i++) 
#define repm(i,n) for(int i = (n)-1; i >= 0; i--)
int N;
int c[10], answer;
int main(void){
	cin >> N;
	int dnum = (int)log10(N) + 1;//Nの桁数

	if(4 <= dnum){
		int i = dnum;
		while(i > 3){//4けた以上の時はその下の桁の最大値を足しておく
			if(i == 4)answer += 6; 
			else answer += 3 * i-1 * 6;
			--i;
		}
	} 

	int tmp = 0;
	int maxNum = (int)pow(10, dnum-1);
	for(int i = N; i >= maxNum; i--){
		tmp = i;
		while(tmp > 0){
			c[tmp % 10]++;	
			tmp /= 10;
		}
		if(!c[0] && !c[1] && !c[2] && !c[4] && !c[6] && !c[8] && !c[9]){ 
			if(c[3] && c[5] && c[7]) answer++;
		}
		memset(c, 0, sizeof(int)*10);
	}
	cout << answer << endl;
	return 0;
}