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)

#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<math.h>

#define rep(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define repi(i,a,b) for(int i=int(a);i<int(b);++i)
#define all(x) (x).begin(),(x).end()

using namespace std;

typedef long long ll;
typedef pair<int, int> P;


template<typename T>
void dump(T *arr, int h, int w){
	rep(i,h){
		rep(j,w) cout << arr[i][j] << "\t";
		cout << "\n";
	}
}

template<typename T>
void dump(T *arr, int a){
	rep(i,a) cout << arr[i] << "\t";
	cout << "\n";
}

template<typename T>
void dump(vector<T> arr, int a){
	rep(i,a) cout << arr[i] << "\t";
	cout << "\n";
}

const int INF = 1e9;
const int MOD = 1e9+7;

bool compare(P a, P b){
	if(a.second != b.second) return a.second > b.second;
	else	return a.first > b.first;
}


int digits(ll num){
	ll d=0;
	while(num >= 1){
		num/= 10;
		d++;
	}
	return d;
}

ll pow_kai(ll a, int n){//aのn乗を計算します。
  ll x = 1;
  while(n > 0){//全てのbitが捨てられるまで。
    if(n&1){//1番右のbitが1のとき。
      x = x*a;
    }
    a = a*a;
    n >>= 1;//bit全体を右に1つシフトして一番右を捨てる。
  }
  return x;
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

	ll n;
	cin >> n;
	
	int depth = log2(n);
	cout << pow_kai(2LL, depth+1)  - 1 << "\n";

	return 0;
}