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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
#define FOR(i,x,n) for(int i=x; i<(n); i++)
#define ALL(n) begin(n),end(n)
#define MOD (1000000007)
#define INF (1e9)
#define INFL (1e18)

typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
template<class T>using arr=vector<vector<T>>;
template<class T>int popcount(T &a){int c=0; rep(i, 8*(int)sizeof(a)){if((a>>i)&1) c++;} return c;}
template<class T>void pr(T x){cout << x << endl;}
template<class T>void prvec(vector<T>& a){rep(i, a.size()-1){cout << a[i] << " ";} pr(a[a.size()-1]);}
template<class T>void prarr(arr<T>& a){rep(i, a.size()) if(a[i].empty()) pr(""); else prvec(a[i]);}
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

ll n;
int ans=0;
int dx[] = {3, 5, 7};

void dfs(ll x, vector<int> sft){
    int d = 1;
    rep(i, 3) d*=sft[i];
    if(d>0) ans++;

    ll t = x;
    rep(i, 3){
        if(10*x + dx[i] <= n){
            vector<int> next = sft;
            next[i]++;
            dfs(10*x + dx[i], next);
        }
    }
}

int main()
{
    cin >> n;
    vector<int> sft(3, 0);
    dfs(0, sft);
    pr(ans);
    return 0;}