Problem:
Score : 500 points

Problem Statement

We have a grid with H horizontal rows and W vertical columns. Let (i,j) denote the square at the i-th row from the top and the j-th column from the left.

The square (i, j) has two numbers A_{ij} and B_{ij} written on it.

First, for each square, Takahashi paints one of the written numbers red and the other blue.

Then, he travels from the square (1, 1) to the square (H, W). In one move, he can move from a square (i, j) to the square (i+1, j) or the square (i, j+1). He must not leave the grid.

Let the unbalancedness be the absolute difference of the sum of red numbers and the sum of blue numbers written on the squares along Takahashi's path, including the squares (1, 1) and (H, W).

Takahashi wants to make the unbalancedness as small as possible by appropriately painting the grid and traveling on it.

Find the minimum unbalancedness possible.

Constraints

2 \leq H \leq 80

2 \leq W \leq 80

0 \leq A_{ij} \leq 80

0 \leq B_{ij} \leq 80

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H W
A_{11} A_{12} \ldots A_{1W}
:
A_{H1} A_{H2} \ldots A_{HW}
B_{11} B_{12} \ldots B_{1W}
:
B_{H1} B_{H2} \ldots B_{HW}

Output

Print the minimum unbalancedness possible.

Sample Input 1

2 2
1 2
3 4
3 4
2 1

Sample Output 1

0

By painting the grid and traveling on it as shown in the figure below, the sum of red numbers and the sum of blue numbers are 3+3+1=7 and 1+2+4=7, respectively, for the unbalancedness of 0.

Sample Input 2

2 3
1 10 80
80 10 1
1 2 3
4 5 6

Sample Output 2

2

Solution in LuaJIT (2.0.4) (judge verdict: Accepted)

local mab = math.abs
local mmi = math.min
local h, w = io.read("*n", "*n")
local a = {}
for i = 1, h do
  for j = 1, w do
    a[(i - 1) * w + j] = io.read("*n")
  end
end
for i = 1, h do
  for j = 1, w do
    local idx = (i - 1) * w + j
    a[idx] = mab(a[idx] - io.read("*n"))
  end
end

-- top
local odd, ev = {}, {}
for i = 1, w do
  odd[i] = {}
  ev[i] = {}
  for j = 1, 6401 do
    odd[i][j] = false
    ev[i][j] = false
  end
end
for i = 1, 81 do odd[1][i] = false end
odd[1][1 + a[1]] = true
for j = 2, w do
  local av = a[j]
  local lim = 80 * j + 1
  for dst = 1, lim do
    local f = false
    if odd[j - 1][1 + mab(dst - 1 - av)] then
      f = true
    end
    if not f and dst + av <= lim - 80 and odd[j - 1][dst + av] then
      f = true
    end
    odd[j][dst] = f
  end
end
local mmi = math.min
-- rem
for i = 2, h do
  local refsrc = i % 2 == 0 and odd or ev
  local refdst = i % 2 == 0 and ev or odd
  do -- j = 1
    local av = a[(i - 1) * w + 1]
    local lim = 80 * i + 1
    for dst = 1, lim do
      local f = false
      if refsrc[1][1 + mab(dst - 1 - av)] then
        f = true
      end
      if not f and dst + av <= lim - 80 and refsrc[1][dst + av] then
        f = true
      end
      refdst[1][dst] = f
    end
  end
  for j = 2, w do
    local av = a[(i - 1) * w + j]
    local lim = mmi(6400, (i + j - 1) * 80) + 1
    for dst = 1, lim do
      local f = false
      if refsrc[j][1 + mab(dst - 1 - av)] or refdst[j - 1][1 + mab(dst - 1 - av)] then
        f = true
      end
      if not f and dst + av <= lim - 80 then
        if refsrc[j][dst + av] or refdst[j - 1][dst + av] then
          f = true
        end
      end
      refdst[j][dst] = f
    end
  end
end
local ret = 1000000007
do
  local lim = (h + w - 1) * 80 + 1
  local refdst = h % 2 == 0 and ev or odd
  for p = 1, lim do
    if refdst[w][p] then
      ret = p - 1
      break
    end
  end
end
print(ret)