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 Python (3.4.3) (judge verdict: Accepted)

from statistics import mean, median,variance,stdev
import numpy as np
import sys
import math
import fractions
import itertools
import copy

def j(q):
    if q==1: print("Heisei")
    else:print("TBD")
    exit(0)


def ct(x,y):
    if (x>y):print("")
    elif (x<y): print("")
    else: print("")

def ip():
    return int(input())


n = ip()                              #入力整数1つ
#h,w= (int(i) for i in input().split())      #入力整数横2つ
#n,x,y = (int(i) for i in input().split())    #入力整数横3つ
#n,a,b,c= (int(i) for i in input().split())  #入力整数横4つ
#a = [int(i) for i in input().split()]        #入力整数配列
#a = input()                                  #入力文字列
#a = input().split()                          #入力文字配列
#n = ip()                      #入力セット(整数改行あり)(1/2)
#a=[ip() for i in range(n)]    #入力セット(整数改行あり)(2/2)

#jの変数はしようできないので注意
#全足しにsum変数使用はsum関数使用できないので注意
x = 1
while x<=n:
    x*=2
print(x//2)