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 <bits/stdc++.h>
#include<iostream>
#include<climits>
#include<math.h>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<stdio.h>
#include <string>
#include <complex>
#include <functional>
#include <map>
#include <queue>
#include <cctype>
using namespace std;
using ll = long long;
const int mod = 1000000007;

ll gcd(ll a, ll b){return b ? gcd(b,a%b) : a;};//最大公約数
ll lcm(ll a, ll b){return a*b/gcd(a,b);};//最小公倍数
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
//typedef pair<string,string> P;
//double dat[100][100];
//int dp[6][1010];//動的計画法
//int prime[100001];
//char str[1010][1010];
//typedef pair<ll,ll> A;
//vector<A> pc;
//int ABS(int a){return max(a,-a);}
//ll a[100000];
//ll dom[100001];

int main(){
  ll n;
  cin>>n;
  ll p = (ll)log2(n);
  ll ans=0;
  //cout<<p;
  while(p){
    ans+=(ll)pow(2,p);
    p--;
  }
  ans++;
  cout<<ans;
   return 0;
}