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 OCaml (4.02.3) (judge verdict: Accepted)

let rec seq a b = if a >= b then [] else a :: seq (a+1) b
let (>>=) x f = List.map f x |> List.concat
let () =
  Scanf.scanf "%d %d" @@ fun n c ->
  let d =
    Array.init c (fun i ->
      Array.init c (fun j -> Scanf.scanf " %d" ((+) 0)))
  in
  let b =
    Array.init n (fun i ->
      Array.init n (fun j -> Scanf.scanf " %d" ((+) 0)))
  in

  let sc = Array.init 3 (fun _ -> Array.make c 0) in
  let _ =
    seq 0 n >>= fun x ->
    seq 0 n >>= fun y ->
      let i = (x + y) mod 3 in
      seq 0 c |> List.iter (fun j -> sc.(i).(j) <- sc.(i).(j) + d.(b.(x).(y)-1).(j));
      []
  in

  (seq 0 c >>= fun i ->
   seq 0 c >>= fun j ->
   seq 0 c >>= fun k ->
     if i = j || j = k || k = i then []
     else [ sc.(0).(i) + sc.(1).(j) + sc.(2).(k) ])
  |> List.fold_left min 10101010101010 
  |> Printf.printf "%d\n"