Problem:
Score : 400 points

Problem Statement

There is a grid with N rows and N columns of squares. Let (i,j) be the square at the i-th row from the top and the j-th column from the left.

These squares have to be painted in one of the C colors from Color 1 to Color C. Initially, (i,j) is painted in Color c_{i,j}.

We say the grid is a good grid when the following condition is met for all i,j,x,y satisfying 1 \leq i,j,x,y \leq N:

If (i+j) \% 3=(x+y) \% 3, the color of (i,j) and the color of (x,y) are the same.

If (i+j) \% 3 \neq (x+y) \% 3, the color of (i,j) and the color of (x,y) are different.

Here, X \% Y represents X modulo Y.

We will repaint zero or more squares so that the grid will be a good grid.

For a square, the wrongness when the color of the square is X before repainting and Y after repainting, is D_{X,Y}.

Find the minimum possible sum of the wrongness of all the squares.

Constraints

1 \leq N \leq 500

3 \leq C \leq 30

1 \leq D_{i,j} \leq 1000 (i \neq j),D_{i,j}=0 (i=j)

1 \leq c_{i,j} \leq C

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N C
D_{1,1} ... D_{1,C}
:
D_{C,1} ... D_{C,C}
c_{1,1} ... c_{1,N}
:
c_{N,1} ... c_{N,N}

Output

If the minimum possible sum of the wrongness of all the squares is x, print x.

Sample Input 1

2 3
0 1 1
1 0 1
1 4 0
1 2
3 3

Sample Output 1

3

Repaint (1,1) to Color 2. The wrongness of (1,1) becomes D_{1,2}=1.

Repaint (1,2) to Color 3. The wrongness of (1,2) becomes D_{2,3}=1.

Repaint (2,2) to Color 1. The wrongness of (2,2) becomes D_{3,1}=1.

In this case, the sum of the wrongness of all the squares is 3.

Note that D_{i,j} \neq D_{j,i} is possible.

Sample Input 2

4 3
0 12 71
81 0 53
14 92 0
1 1 2 1
2 1 1 2
2 2 1 3
1 1 2 2

Sample Output 2

428

Solution in Lua (5.3.2) (judge verdict: Accepted)

local mmin, mfl, ior = math.min, math.floor, io.read
local n, c = ior("*n", "*n")
local d = {}
for i = 1, c * c do
  d[i] = ior("*n")
end
local diffsum = {}
for i = 1, 3 * c do
  diffsum[i] = 0
end
for pos = 1, n * n do
  local row, col = 1 + mfl((pos - 1) / n), pos % n
  if(col == 0) then col = n end
  local md = (row + col) % 3
  local a = ior("*n")
  for i_c = 1, c do
    local idx = md * c + i_c
    diffsum[idx] = diffsum[idx] + d[(a - 1) * c + i_c]
  end
end
local dfmin = diffsum[1] + diffsum[c + 2] + diffsum[2 * c + 3]
for i_1 = 1, c do
  for i_2 = 1, c do
    if(i_1 ~= i_2) then
      for i_3 = 1, c do
        if(i_1 ~= i_3 and i_2 ~= i_3) then
          dfmin = mmin(dfmin, diffsum[i_1] + diffsum[c + i_2] + diffsum[2 * c + i_3])
        end
      end
    end
  end
end
print(dfmin)