output_description: Print two integers x and y separated by a space. If there are several pairs of
  such x and y, print that pair for which |x| + |y| is the minimal (primarily)
  and x ≤ y (secondarily).
input_description: a b
      
  
  Two positive integers a and b are given separated by a space in a line.
attempt: # -*- coding: utf-8 -*-
  
  import sys
  
  
  def input():
      return sys.stdin.readline().strip()
  
  
  def list2d(a, b, c):
      return [[c] * b for i in range(a)]
  
  
  def list3d(a, b, c, d):
      return [[[d] * c for j in range(b)] for i in range(a)]
  
  
  def list4d(a, b, c, d, e):
      return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
  
  
  def ceil(x, y=1):
      return int(-(-x // y))
  
  
  def INT():
      return int(input())
  
  
  def MAP():
      return map(int, input().split())
  
  
  def LIST(N=None):
      return list(MAP()) if N is None else [INT() for i in range(N)]
  
  
  def Yes():
      print("Yes")
  
  
  def No():
      print("No")
  
  
  def YES():
      print("YES")
  
  
  def NO():
      print("NO")
  
  
  sys.setrecursionlimit(10**9)
  INF = float("inf")
  MOD = 10**9 + 7
  
  
  # 拡張ユークリッドの互除法(ax+by=gcd(a, b)の解を求める)
  def extgcd(a, b, x, y):
      if b == 0:
          x = 1
          y = 0
          return (y, x)
      else:
          x, y = extgcd(b, a % b, y, x)
          y -= a // b * x
          return (y, x)
  
  
  a, b = MAP()
  
  y, x = extgcd(a, b, 0, 0)
  print(x, y)
  
problem_description: Extended Euclid Algorithm
  
    
  
  Given positive integers a and b, find the integer solution (x, y) to ax + by =
  gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.
samples: [{"input": "4 12", "output": "1 0"}, {"input": "3 8", "output": "3 -1"}]