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)

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
 
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
 
using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
 
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
 
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
 
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
 
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
 
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001; //check the limits, dummy
 
int main() {
	ios_base::sync_with_stdio(0); cin.tie(0);    
	
    int cnt = 0;
    ll N; cin >> N;
    while (N > 0) {
        N /= 2; cnt++;
    }

    cout << (1ll << cnt) - 1 << endl;
	
	return 0;
}
 
// read the question correctly (ll vs int)
// template by bqi343