You are given an integer N. Consider all possible segments on the coordinate axis with endpoints at integer points with coordinates between 0 and N, inclusive; there will be <image> of them.

You want to draw these segments in several layers so that in each layer the segments don't overlap (they might touch at the endpoints though). You can not move the segments to a different location on the coordinate axis. 

Find the minimal number of layers you have to use for the given N.

Input

The only input line contains a single integer N (1 ≤ N ≤ 100).

Output

Output a single integer - the minimal number of layers required to draw the segments for the given N.

Examples

Input

2


Output

2


Input

3


Output

4


Input

4


Output

6

Note

As an example, here are the segments and their optimal arrangement into layers for N = 4.

<image>
'''
a lower bound is given by the number of intervals covering z = n/2-epsilon.
each such interval is given by endpoints x,y where x\in [0,n/2) and y\in[n/2,n].
if n=2m is even,  then the number of ways to choose endpoints is m*(m+1) and
if n=2m+1 is odd, then the number of ways to choose endpoints is (m+1)^2.

is the minimum attained? the greedy algorithm (see below) works for all 1<=n<=100 but i haven't thought this through entirely yet
'''

from collections import defaultdict

def lower(n):
	if n % 2 == 0:
		m = n//2
		return (m*(m+1))
	if n % 2 == 1:
		m = n//2
		return ((m+1)**2)

def layers(n, verbose=True):
	'''create layers using greedy algorithm:
	always choose the remaining interval [x,y] with the least x and subsequently largest y.
	this seems to always produce only layers with at most two intervals, and the number of layers matches 
	'''

	# first create all intervals
	d = defaultdict(list) # d[x] is the set of intervals [x,y] for some y
	num_intervals = 0
	for x in range(n+1):
		for y in range(x+1,n+1):
			d[x].append( (x,y) )
			num_intervals += 1

	# next create layers greedily
	num_layers = 0
	while num_intervals > 0:
		num_layers += 1

		# create a layer:
		i = 0

		while i <= n:
			if len(d[i]) > 0:
				x,y = d[i].pop() # take an interval
				num_intervals -= 1
				if verbose:
					print (x,y),
				i = y
			else:
				i += 1
		if verbose:
			print

	if num_layers != lower(n):
		print 'discrepancy, lower=', lower(n)
	return num_layers

# # testing that it works:
# for n in range(1,101):
# 	print layers(n,False), lower(n)

n = int(raw_input())
print lower(n)