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)

#include <bits/stdc++.h>
#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
using i64 = int64_t;
using f80 = __float80;
static const int dy[4] = {0,1,0,-1}, dx[4] = {1,0,-1,0};
constexpr i64 INF = 1'010'000'000'000'000'017LL;
constexpr i64 MOD = 1'000'000'007LL;
#define _overload3(_1,_2,_3,name,...) name
#define repi(i,start,end) for(i64 i=(start);i<(end);++i)
#define _rep(i,n) repi(i,0,n)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define rrep(i,start,end) for(i64 i=(start);i>end;--i)
#define rng(a) a.begin(),a.end()
#define rrng(a) a.rbegin(),a.rend()
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,v(T),greater<T> >
#define v(T) vector<T>
#define vv(T) v(v(T))
typedef v(i64) vi;
typedef v(vi) vvi;
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return true;}return false;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return true;}return false;}
template<typename T> ostream &operator<<(ostream &os, const vector< T > &v){
  rep(i,v.size()) os << v[i] << (i + 1 != v.size() ? " " : "");
  return os;
}
template<typename T> istream &operator>>(istream &is, vector< T > &v){
  for(T &in : v) is >> in;
  return is;
}

void Main()
{
  int N;
  cin >> N;
  int i = 1;
  while((i<<1)<=N) i=i<<1;
  cout << i << "\n";

}

int main()
{
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(15);
  Main();
}