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 (Clang 10.0.0) (judge verdict: Accepted)

#include <stdio.h>
#include <math.h>

int main(void) {
  int input;
  fscanf(stdin, "%d", &input);

  int tmp;
  // 2の累乗に最も近い値を選択
  for (int i = input; i > 0; i--) {
    if (log2(i) == 0.0 // 1
        || log2(i) == 1.0 // 2
        || log2(i) == 2.0 // 4
        || log2(i) == 3.0 // 8
        || log2(i) == 4.0 // 16
        || log2(i) == 5.0 // 32
        || log2(i) == 6.0 // 64
        ) {
      tmp = i;
      break;
    }
  }

  // tmp以上input以下を走査
  int val = 0; // 割った回数
  int output = tmp; // 最終的な出力値
  for (int i = tmp; i <= input; i++) {
    int iternum = 0;
    int n = i;
    for(;;) {
      if (n % 2 == 0) {
        iternum++;
        n = n / 2;
      } else {
        break;
      }
    }

    if (val <= iternum) {
      val = iternum;
      output = i;
    }
  }

  printf("%d\n", output);
  return 0;
}