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: Accepted)

// --------------------<optimizations>--------------------
#pragma GCC optimize("O3")
//(UNCOMMENT WHEN HAVING LOTS OF RECURSIONS)\
#pragma comment(linker, "/stack:200000000")
//(UNCOMMENT WHEN TRYING TO BRUTEFORCE WITH A LOT OF LOOPS)\
#pragma GCC optimize("unroll-loops")

#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")

#include<bits/stdc++.h>

using namespace std;

#define int long long
#define ll long long
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define pii pair<ll,ll>
#define vpii vector<pair<ll,ll> >
#define F first
#define S second
#define ld long double
#define built __builtin_popcountll
#define mst(a,i) memset(a,i,sizeof(a))
#define all(x) x.begin(),x.end()
#define itit(it,a) for(auto it=(a).begin(); it!=(a).end(); it++)
#define rep(i,a,b) for(ll i=a; i<b; i++)
#define repr(i,a,b) for(ll i=a; i>b; i--)
#define reprr(i,a,b) for(ll i=a; i>=b; i--)
#define pi 3.14159265358979323846264338327950288419716939937510582097494459230
ll max3(ll x,ll y,ll z) {return max(max(x,y),z);}
ll min3(ll x,ll y,ll z) {return min(min(x,y),z);}
const ll N=1e5+10,M=2e5+10,M2=1e6+10,mod=1e9+7,inf=1e17+10;
const int INF=1e9+7;
void add(int& a, int b) {a+=b;if(a>=mod){a-=mod;}}

#define trace(...) cerr<<"Line:"<<__LINE__<<" "; __f(#__VA_ARGS__, __VA_ARGS__)
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
    cerr<<name<<" : "<<arg1<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
    const char *comma=strchr(names+1,',');
    cerr.write(names,comma-names)<<" : "<<arg1<<" | ";
    __f(comma+1,args...);
}

int X[] = {0, 1, 0, -1};
int Y[] = {-1, 0, 1, 0};
//assic value of ('0'-'9') is(48 - 57) and (a-z) is (97-122) and (A-Z) is(65-90) and 32 for space
ll power(ll x,ll n){ll result=1;while(n>0){if(n%2==1)result=(result*x)%mod; x=((x%mod)*(x%mod))%mod;n=n/2;}return result;}


int32_t main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ll h;
    cin>> h;
    ll ans=0;
    int j=1;
    while(h){
        ans+=j;
        h/=2;
        j*=2;
    }
    cout << ans;
    return 0;
}