Problem:
Score : 400 points

Problem Statement

Caracal is fighting with a monster.

The health of the monster is H.

Caracal can attack by choosing one monster. When a monster is attacked, depending on that monster's health, the following happens:

If the monster's health is 1, it drops to 0.

If the monster's health, X, is greater than 1, that monster disappears. Then, two new monsters appear, each with the health of \lfloor X/2 \rfloor.

(\lfloor r \rfloor denotes the greatest integer not exceeding r.)

Caracal wins when the healths of all existing monsters become 0 or below.

Find the minimum number of attacks Caracal needs to make before winning.

Constraints

1 \leq H \leq 10^{12}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H

Output

Find the minimum number of attacks Caracal needs to make before winning.

Sample Input 1

2

Sample Output 1

3

When Caracal attacks the initial monster, it disappears, and two monsters appear, each with the health of 1.

Then, Caracal can attack each of these new monsters once and win with a total of three attacks.

Sample Input 2

4

Sample Output 2

7

Sample Input 3

1000000000000

Sample Output 3

1099511627775

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

/*
in the name of ALLAH
author: Sepehr_Noori
*/
#include <bits/stdc++.h>

using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define pp pop_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef long long int ll;
typedef map<int,int> mpii;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ll,int> pli;
typedef pair<int,ll> pil;
typedef double db;
typedef long double ld;
const int mod = 1e9+7;
////////////////////////
int solve(ll h){
    if(h == 1){
        // s++;
        return 1;
    }
    ll s=0,g=0;
    if(h%2 != 0){
        h--;
    }
    h/=2;
    s++;
    g = solve(h);
    s += 2*g;
    cout<<h<<" "<<s<<" "<<g<<endl;
    return s;
}

int main(){
    ios::sync_with_stdio(false);cin.tie(0);
    ll h;
    cin>>h;
    ll ans = solve(h); 
    cout<<ans<<endl;
}
    /*ll s=0,k,n,x;
    vi h;
    cin>>n>>k;
    for (int i =1;i<=n;i++){
        cin>>x;
        h.pb(x);
    }
    sort(all(h));
    while(k-- &&SZ(h))
        h.pp();
    for (auto z:h){
        s+=z;
    }
    cout<<s<<endl;

}
    /*int h,n,x,s=0;
    cin>>h>>n; 
    for (int i = 0;i<n;i++){
        cin>>x;
        s+=x;
    }
    if(s>=h){
        cout<<"Yes"<<endl;
        return 0;
    }else
    {
        cout<<"No"<<endl;
        return 0;
    }
    
}
    /*db h,a;
    cin>>h>>a;
    cout<<ceil(h/a)<<endl;;

    return 0;
}*/