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)

import math
from collections import deque
from collections import defaultdict

def readInt():
	return int(input())
def readInts():
	return list(map(int, input().split()))
def readChar():
	return input()
def readChars():
	return input().split()
def factorization(n):
	res = []
	if n%2==0:
		res.append(2)
	for i in range(3,math.floor(n//2)+1,2):
		if n%i==0:
			c = 0
			for j in res:
				if i%j==0:
					c=1
			if c==0:
				res.append(i)
	return res
def fact2(n):
	p = factorization(n)
	res = []
	for i in p:
		c=0
		z=n
		while 1:
			if z%i==0:
				c+=1
				z/=i
			else:
				break
		res.append([i,c])
	return res
def fact(n):#階乗
	ans = 1
	m=n
	for _i in range(n-1):
		ans*=m
		m-=1
	return ans
def comb(n,r):#コンビネーション
	if n<r:
		return 0
	l = min(r,n-r)
	m=n
	u=1
	for _i in range(l):
		u*=m
		m-=1
	return u//fact(l)
def combmod(n,r,mod):
	return (fact(n)/fact(n-r)*pow(fact(r),mod-2,mod))%mod
def printQueue(q):
	r=copyQueue(q)
	ans=[0]*r.qsize()
	for i in range(r.qsize()-1,-1,-1):
		ans[i] = r.get()
	print(ans)
class UnionFind():
	def __init__(self, n):
		self.n = n
		self.parents = [-1]*n

	def find(self, x): # root
		if self.parents[x]<0:
			return x
		else:
			self.parents[x] = self.find(self.parents[x])
			return self.parents[x]

	def union(self,x,y):
		x = self.find(x)
		y = self.find(y)

		if x==y:
			return

		if self.parents[x]>self.parents[y]:
			x,y = y,x

		self.parents[x]+=self.parents[y]
		self.parents[y]=x

	def size(self,x):
		return -1*self.parents[self.find(x)]

	def same(self,x,y):
		return self.find(x)==self.find(y)

	def members(self,x): # much time
		root = self.find(x)
		return [i for i in range(self.n) if self.find(i)==root]

	def roots(self):
		return [i for i,x in enumerate(self.parents) if x<0]

	def group_count(self):
		return len(self.roots())

	def all_group_members(self):
		return {r: self.members(r) for r in self.roots()} # 1~n
def bitArr(n):#ビット全探索
	x = 1
	zero = "0"*n
	ans = []
	ans.append([0]*n)
	for i in range(2**n-1):
		ans.append(list(map(lambda x:int(x),list((zero+bin(x)[2:])[-1*n:]))))
		x+=1
	return ans;
def arrsSum(a1,a2):
	for i in range(len(a1)):
		a1[i]+=a2[i]
	return a1
def maxValue(a,b,v):
	v2 = v
	for i in range(v2,-1,-1):
		for j in range(v2//a+1): #j:aの個数
			k = i-a*j
			if k%b==0:
				return i
	return -1
def copyQueue(q):
	nq = queue.Queue()
	n = q.qsize()
	for i in range(n):
		x = q.get()
		q.put(x)
		nq.put(x)
	return nq

n = readInt()

ans=1

while ans*2<=n:
	ans*=2

print(ans)