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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <iostream>
#include <iostream>
#include <fstream>
#include <numeric>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <cmath>
#include <set>
#include <map>
#include <functional>
#include <bitset>
#include <iomanip>
#include <stack>
#include <list>
#include <cstdint>
using namespace std;
#define     loop(n,i,a)     for(ll i=a;i<n;i++)
#define     loopR(n,i,a)    for(ll i=a-1;i>=n;i--)
#define     all(arr,n)      arr,arr+n
#define     allv(v)         (v).begin(),(v).end()
#define     rallv(v)        (v).rbegin(),(v).rend()
#define     m_p             make_pair
#define     ll              long long
#define     pii             pair<ll,ll>
#define     vi              vector<int>
#define     vll             vector<ll>
#define     vii             vector<pair<ll,ll>>
#define     sz(x)           (int)x.size()
#define     pb              push_back
#define     endl            "\n"
#define     Endl            "\n"
#define     f               first
#define     s               second
#define     mem(dp,n)       memset(dp,n,sizeof dp)
int dx[] = { 1 , 0 ,-1 , 0 ,-1 ,-1 , 1 , 1 };
int dy[] = { 0 , 1 , 0 ,-1 ,-1 , 1 ,-1 , 1 };
int KnightI[] = { 2, 1, -1, -2, -2, -1, 1, 2 };
int KnightJ[] = { 1, 2, 2, 1, -1, -2, -2, -1 };
template<typename T>
void max_self(T &a,T b){
    a=max(a,b);
}
template<typename T>
void min_self(T &a,T b){
    a=min(a,b);
}
void fast(){
    std::ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
}
vector<string> vec_splitter(string s) {
    s += ',';
    vector<string> res;
    while(!s.empty()) {
        res.push_back(s.substr(0, s.find(',')));
        s = s.substr(s.find(',') + 1);
    }
    return res;
}
void debug_out(
        vector<string> __attribute__ ((unused)) args,
        __attribute__ ((unused)) int idx,
        __attribute__ ((unused)) int LINE_NUM) { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
    if(idx > 0) cerr << ", "; else cerr << "Line(" << LINE_NUM << ") ";
    stringstream ss; ss << H;
    cerr << args[idx] << " = " << ss.str();
    debug_out(args, idx + 1, LINE_NUM, T...);
}
#define debug(...) debug_out(vec_splitter(#__VA_ARGS__), 0, __LINE__, __VA_ARGS__)
const ll mxN=5e5+10,oo=0x3f3f3f3f,MOD=1e9+7;
const double PI = acos(-1);

void solve(){
    int n;
    cin>>n;
    int res=1,mx=0;
    loop(n,i,2){
        int tmp=i;
        int cnt=0;
        while(tmp%2==0)
            cnt++,tmp/=2;
        if(cnt>mx)
            res=i,mx=cnt;
    }
    cout<<res<<Endl;
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    fast();
    solve();

}