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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define ABS(a) ((a)>(0)?(a):-(a))
#define rep(i,a,b) for(int (i)=int(a);(i)<int(b);(i)++)
#define rrep(i,a,b) for(int (i)=int(a);(i)>=int(b);(i)--)

#define put(a) cout << (a) << endl
#define puts(a) cout << (a) << " "

#define llint long long int

#define INF (long long)(1e9+1)
#define MOD (long long)(1e9+7)
#define INF64 (long long)(1e18+1)

#define F first
#define S second

#define Pii  pair<int,int>
#define Pll  pair<long long,long long>
#define Piii pair<int,pair<int,int>>
#define Plll pair<long long,pair<long long,long long>>

#define Vll(a,b,c)    vector<vector<long long>> (a)((b),vector<long long>((c))
#define Vlll(a,b,c,d) vector<vector<vector<long long>>> (a)((b),vector<vector<long long>>((c),vector<long long>((d)))

#define MAX_N 1000000

llint my_binary_search(llint ary[], llint key, llint left, llint right) {
	llint middle;
	if (right < left) {
		return -(INF64);
	} else {
		middle = left + (right - left) / 2;
		if (ary[middle] > key) {
			return my_binary_search(ary, key, left, middle - 1);
		} else if (ary[middle] < key) {
			return my_binary_search(ary, key, middle + 1, right);
		} else {
			return middle;
		}
	}
}

struct UnionFind {
	vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

	UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
		for(int i = 0; i < N; i++) par[i] = i;
	}

	int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
		if (par[x] == x) return x;
		return par[x] = root(par[x]);
	}

	void unite(int x, int y) { // xとyの木を併合
		int rx = root(x); //xの根をrx
		int ry = root(y); //yの根をry
		if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
		par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
	}

	bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
		int rx = root(x);
		int ry = root(y);
		return rx == ry;
	}
};

llint dps(llint h) {
	if (h==1) {
		return 1;
	} else {

		return 1+2*dps(h/2);
	}
}

int main (void)
{
	//INPUT
	llint H;
	cin>> H;

	llint ans=dps(H);

	cout << ans << endl;

	return 0;
}