Problem:
Score : 200 points

Problem Statement

Takahashi loves numbers divisible by 2.

You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique.

Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder.

For example,

6 can be divided by 2 once: 6 -> 3.

8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1.

3 can be divided by 2 zero times.

Constraints

1 ≤ N ≤ 100

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

7

Sample Output 1

4

4 can be divided by 2 twice, which is the most number of times among 1, 2, ..., 7.

Sample Input 2

32

Sample Output 2

32

Sample Input 3

1

Sample Output 3

1

Sample Input 4

100

Sample Output 4

64

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

// lcmとかはg++ -std=c++17 default.cppみたいなかんじで
//vector<vector<ll>> piyo(tate,vector<ll> (yoko));

#include <bits/stdc++.h>
#define mod 1000000007
#define INF 1000000000
#define ll long long
#define ln cout<<endl;
#define Yes cout<<"Yes"<<endl;
#define No cout<<"No"<<endl;
#define double long double
#define REP(i,m,n) for(ll i=(ll)(m);i<(ll)(n);i++)
#define rep(i,n) REP(i,0,n)
using namespace std;

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

    ll a,b,c,d,m,n,maxi=0,f=0,mini=INF,sum=0;
    string str1,str2;
    cin>>n;
    while(n>=2){
        n/=2;
        sum++;
    }
    a=1;
    rep(i,sum){
        a=a*2;
    }
    cout<<a<<endl;
    return 0;
}