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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun main(args: Array<String>) {
    val (N, C) = readLine()!!.split(" ").map{it.toInt()}

    val D = Array(C) { readLine()!!.split( " ").map{it.toInt()} }
    val c = Array(N) { readLine()!!.split(" ").map{it.toInt()} }

    val a = Array(3) { LongArray(C) { 0L } }
    val b = Array(3) { 0L }

    for (i in c.indices) for (j in c[i].indices) {
        val k = (i + j + 2) % 3

        a[k][c[i][j] - 1] += 1L
        b[k] += 1L
    }

    var res = 100000000L
    for (c1 in 0 until C) for (c2 in D.indices) for (c3 in D.indices) {
        if (c1 == c2 || c2 == c3 || c3 == c1) continue

        var s = 0L
        val cs = arrayOf(c1, c2, c3)
        for (i in a.indices) {
            for (j in a[i].indices) {
                if (j == cs[i]) continue

                s += a[i][j] * D[j][cs[i]]
            }
        }

        res = Math.min(res, s)
    }

    println(res)
}