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: Accepted)

#include <bits/stdc++.h>
typedef long long int ll;
typedef long double ld;
using namespace std;
const long long int INF = 1e18;
const long long int mod = 1e9+7;

typedef pair<ll, ll> pairs;
typedef vector<pairs> p;

struct Edge{
    ll to, weight;
    Edge(ll t, ll w) : to(t), weight(w){}
};
using graph = vector<vector<ll>>;
using Graph = vector<vector<Edge>>;


ll gcd(ll a,ll b){
    if(b == 0){
        return a;
    }else{
        return gcd(b, a%b);
    }
}

ll lcm(ll a, ll b){
    return a / gcd(a, b) * b;
}

ll keta(ll N){
    int tmp{};
    while( N > 0 ){
        tmp += ( N % 10 );
        N /= 10;
    }
    N = tmp;
    return N;
}

// 回文
bool kai(string S){
    bool flag = true;
    for(ll i=0;i<S.size()/2;++i){
        if(S[i] != S[S.size()-i-1]){
            flag = false;
            break;
        }
    }
    return flag;
}
// ---------------------------------------------

ll ans = 0;
ll n;
void dfs(ll x, bool three, bool five, bool seven){
    if(x > n){
        return;
    }
    if(three and five and seven){
        ans++;
    }
    dfs(10*x+3, true, five, seven);
    dfs(10*x+5, three, true, seven);
    dfs(10*x+7, three, five, true);
}

int main(){
    cin>>n;
    dfs(0, false, false, false);
    cout<<ans<<endl;
}