output_description: Print the minimum possible value of S_{max} - S_{min}.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      H W
attempt: # -*- coding: utf-8 -*-
  import sys
  import copy
  import collections
  from bisect import bisect_left
  from bisect import bisect_right
  from collections import defaultdict
  from heapq import heappop, heappush
  import math
  import itertools
  import random
   
  # NO, PAY-PAY
  #import numpy as np
  #import statistics
  #from statistics import mean, median,variance,stdev
   
  def inputInt(): return int(input())
  def inputMap(): return map(int, input().split())
  def inputList(): return list(map(int, input().split()))
   
  def main():
      H,W = inputMap()
      
      if H % 3 == 0 or W % 3 == 0:
          print(0)
          sys.exit()
          
      if H <= 2 and W <= 2:
          print(1)
          sys.exit()
          
      if H <= W:
          a = -(-W // 3)
          b = -(-H // 2)
          s1 = a * H
          s2 = (W-a) * b
          s3 = (W-a) * (H-b)
          ans1 = max(s1,s2,s3)-min(s1,s2,s3)
          
          a -= 1
          s1 = a * H
          s2 = (W-a) * b
          s3 = (W-a) * (H-b)
          ans2 = max(s1,s2,s3)-min(s1,s2,s3)
          
          a += 2
          s1 = a * H
          s2 = (W-a) * b
          s3 = (W-a) * (H-b)
          ans3 = max(s1,s2,s3)-min(s1,s2,s3)
          
          ans = min(ans1,ans2,ans3))
          
          tmp = W // 3
          tmp2 = (W - tmp) // 2
          s1 = tmp * H
          s2 = tmp2 * H
          s3 = (W - tmp- tmp2) * H
          ans1 = max(s1,s2,s3)-min(s1,s2,s3)
          
          tmp += 1
          s1 = tmp * H
          s2 = tmp2 * H
          s3 = (W - tmp- tmp2) * H
          ans2 = max(s1,s2,s3)-min(s1,s2,s3)
          
          tmp -= 2
          s1 = tmp * H
          s2 = tmp2 * H
          s3 = (W - tmp- tmp2) * H
          ans3 = max(s1,s2,s3)-min(s1,s2,s3)
          
          print(min(ans,ans1,ans2,ans3))
          
      else:
          a = -(-H // 3)
          b = -(-W // 2)
          s1 = a * W
          s2 = (H-a) * b
          s3 = (H-a) * (W-b)
          ans1 = max(s1,s2,s3)-min(s1,s2,s3)
          
          a -= 1
          s1 = a * W
          s2 = (H-a) * b
          s3 = (H-a) * (W-b)
          ans2 = max(s1,s2,s3)-min(s1,s2,s3)
          
          a += 2
          s1 = a * W
          s2 = (H-a) * b
          s3 = (H-a) * (W-b)
          ans3 = max(s1,s2,s3)-min(s1,s2,s3)
          
          ans = min(ans1,ans2,ans3)
          
          tmp = H // 3
          tmp2 = (H - tmp) // 2
          s1 = tmp * W
          s2 = tmp2 * W
          s3 = (H - tmp- tmp2) * W
          ans1 = max(s1,s2,s3)-min(s1,s2,s3)
          
          tmp += 1
          s1 = tmp * W
          s2 = tmp2 * W
          s3 = (H - tmp- tmp2) * W
          ans2 = max(s1,s2,s3)-min(s1,s2,s3)
          
          tmp -= 2
          s1 = tmp * W
          s2 = tmp2 * W
          s3 = (H - tmp- tmp2) * W
          ans3 = max(s1,s2,s3)-min(s1,s2,s3)
          
          print(min(ans,ans1,ans2,ans3))
  	
  if __name__ == "__main__":
  	main()
  
problem_description: Statement
  
  There is a bar of chocolate with a height of H blocks and a width of W blocks.
  Snuke is dividing this bar into exactly three pieces. He can only cut the bar
  along borders of blocks, and the shape of each piece must be a rectangle.
  
  Snuke is trying to divide the bar as evenly as possible. More specifically, he
  is trying to minimize S_{max} \- S_{min}, where S_{max} is the area (the
  number of blocks contained) of the largest piece, and S_{min} is the area of
  the smallest piece. Find the minimum possible value of S_{max} - S_{min}.
samples: [{"input": "3 5", "output": "0\n    \n\nIn the division below, S_{max} - S_{min} = 5 - 5 = 0.\n\n![2a9b2ef47b750c0b7ba3e865d4fb4203.png](https://atcoder.jp/img/arc074/2a9b2ef47b750c0b7ba3e865d4fb4203.png)\n\n* * *"}, {"input": "4 5", "output": "2\n    \n\nIn the division below, S_{max} - S_{min} = 8 - 6 = 2.\n\n![a42aae7aaaadc4640ac5cdf88684d913.png](https://atcoder.jp/img/arc074/a42aae7aaaadc4640ac5cdf88684d913.png)\n\n* * *"}, {"input": "5 5", "output": "4\n    \n\nIn the division below, S_{max} - S_{min} = 10 - 6 = 4.\n\n![eb0ad0cb3185b7ae418e21c472ff7f26.png](https://atcoder.jp/img/arc074/eb0ad0cb3185b7ae418e21c472ff7f26.png)\n\n* * *"}, {"input": "100000 2", "output": "1\n    \n\n* * *"}, {"input": "100000 100000", "output": "50000"}]