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: Wrong Answer)

#include<bits/stdc++.h>
#define fi first
#define se second
#define rep(i, n) for(int (i) = 0; (i) < (n); (i)++)
#define rrep(i, n) for (int (i) = 1; (i) <= (n); (i++))
#define rng(a) a.begin(),a.end()
#define rrng(a) a.rbegin(),a.rend()
#define pb push_back
#define eb emplace_back
#define yn {puts("Yes");}else {puts("No");}
#define printd(x) printf("%.12f\n", x);
template<class T> bool chmin(T& a, const T& b) {if (a > b) {a = b; return 1;} return 0;}
template<class T> bool chmax(T& a, const T& b) {if (a < b) {a = b; return 1;} return 0;}
using namespace std;
using ll = long long;

int main() {
	int n;
	cin >> n;

	int ans = 1;
	int A;
	for (int i = 1; i <= n; i++) {
		int cnt = 0;
		int t = i;
		while (t % 2 == 0) {
			t /= 2;
			cnt++;
		}
		if (chmax(ans, cnt)) A = i;
	}

	cout << A << endl;
	return 0;
}