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)

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define pp pair<pair<ll, ll>,pair<ll, ll>>
#define pll pair<ll,ll>
#define pdd pair<double,double>
#define vii vector<int>
#define vll vector<ll>
#define mat vector<vector<ll>>
#define lb lower_bound
#define ub upper_bound
#define pb push_back
#define fi first
// #define sc second
#define rep(i,n) for(ll i=0;i<n;i++)
#define rep2(i,a,b) for(ll i=a;i<b;i++)
#define repr(i,n) for(ll i=n-1;i>=0;i--)
#define all(x) x.begin(),x.end()
#define sz(x) (ll) (x).size()
#define pq priority_queue<ll>
#define pqg priority_queue<ll,vector<ll>,greater<ll>>
#define LB(v,x) (lower_bound(v.begin(),v.end(),x)-v.begin())
#define UB(v,x) (upper_bound(v.begin(),v.end(),x)-v.begin())
#define ERASE(v) sort(v.begin(),v.end());v.erase(unique(v.begin(),v.end()),v.end())
#define int ll
using namespace std;
const ll INF = (1 << 30 ) - 1;
const ll LLINF = (1LL << 60LL);
const ll MOD = 1000000007;
const ll mod = 998244353;
const ll MAX = 1100000;
const double pi = acos(-1);
const double eps = 1e-10;
ll dx[4] ={1,0,-1,0} , dy[4] ={0,1,0,-1};

template<class T> 
inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> 
inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

struct Timer{
    chrono::system_clock::time_point start, end;
    Timer(){ start = chrono::system_clock::now(); }
    ~Timer(){
        end = chrono::system_clock::now();
        auto msec = chrono::duration_cast<chrono::milliseconds>(end - start).count();
        cerr<<"time : "<<msec<<" ms"<<endl;
    }
};


ll n;
vll v;
map<ll,ll> mp;
void dfs(ll x){
    if(mp[x]>0) return ;
    mp[x]++;
    if(x>1000000000) return ;
    string s=to_string(x);
    ll m=s.size();
    v.pb(x*10+3);dfs(x*10+3);
    v.pb(x*10+5);dfs(x*10+5);
    v.pb(x*10+7);dfs(x*10+7);
    // v.pb(3*(m+1)+x);dfs(3*(m+1)+x);
    // v.pb(5*(m+1)+x);dfs(5*(m+1)+x);
    // v.pb(7*(m+1)+x);dfs(7*(m+1)+x);
    return ;
}


signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    // Timer TM;
    
    cin>>n;
    dfs(0);
    ERASE(v);
    ll m=v.size();
    vll k;
    rep(i,m){
        string s=to_string(v[i]);
        // if(i<100)cout<<s<<endl;
        ll cnt[4]={};
        rep(j,(ll)s.size()){
            if(s[j]-'0'==3) cnt[0]++;
            if(s[j]-'0'==5) cnt[1]++;
            if(s[j]-'0'==7) cnt[2]++;
        }
        if(cnt[0]>0&&cnt[1]>0&&cnt[2]>0){
            k.pb(v[i]);
            // if(i<100) cout<<v[i]<<endl;
        }
    }
    ll s=LB(k,n);
    // cout<<k[s]<<endl;
    cout<<s<<endl;
    return 0;
}