A + B is often used as an example of the easiest problem possible to show some contest platform. However, some scientists have observed that sometimes this problem is not so easy to get accepted. Want to try?
The input contains two integers *a* and *b* (0 ≤ *a*, *b* ≤ 10<sup class="upper-index">3</sup>), separated by a single space.
from math import ceil
	
def mod_expo(n, p, m):
	"""find (n^p)%m"""
	result = 1
	while p != 0:
		if p%2 == 1:
			result = (result * n)%m
		p //= 2
		n = (n * n)%m
	return result
	
def find_product(n, m, arr):
	"""find |arr i - arr j| mod m"""
	product=1
	if n <= m:
		for i in range(n):
			for j in range(i+1, n):
				 product = (product * abs(arr[i] - arr[j]))%m
		return product
	return 0

def count_tables(l):
	count=0
	l.sort()
	a = l[1]-l[0]
	b = l[2]-l[0]
	if a%3 > 0 and (a%3 + (b - (a - a%3))) >= 3:
		count += 1
	count += l[0] + (2*a)//3
	return int(min(sum(l)/3, l[0] + l[1]))
	
def no_vacancy(n, arr):
	hotel = {}
	for i in range(n):
		room = (i + arr[i])%n
		if room in hotel:
			return False
		else:
			hotel[room] = 1
	return True
	

t = 1
#t = int(input())
while t:
	t = t - 1
	#n = int(input())
	# a, b, m = map(int, input().split())
	a, b = map(int, input().split())
	#l = list(map(int, input().strip().split()))[:n]
	#str1 = input()
	#if no_vacancy(n, l):
	#	print("Yes")
	#else:
	#	print("No")
	print(a + b)
	
