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 Ruby (2.3.3) (judge verdict: Accepted)

# Problem https://atcoder.jp/contests/abc068/tasks/abc068_b
# Ruby 2nd Try
Nmax = 100
Ni = Array.new(Nmax+1,0)
1.upto(Nmax) do |j|
    number = j
    counter = 0
    oddFlag = false
    while true
        if number % 2 == 1
        oddFlag = true
        end
        if oddFlag
            break
        end
        number = number / 2
        counter = counter +1
    end
    Ni[j]=counter
end
# p Ni
N = gets.chomp.to_i
answer = 0
countents = 0
1.upto(N) do |j|
    if(countents <= Ni[j])
        answer = j
        countents = Ni[j]
    end
end
puts(answer)