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: Accepted)

fun main(args: Array<String>) {
    val (N, C) = readLine()!!.split(" ").map { it.toInt() }
    val D = mutableListOf<IntArray>()
    val c = mutableListOf<IntArray>()
    (1..C).forEach {
        val d = readLine()!!.split(" ").map { it.toInt() }
        D.add(d.toIntArray())
    }
    (1..N).forEach {
        val ci = readLine()!!.split(" ").map { it.toInt() }
        c.add(ci.toIntArray())
    }




    val zeroMod3Scores = IntArray(C)
    val oneMod3Scores = IntArray(C)
    val twoMod3Scores = IntArray(C)

    // O(CN^2)
    (1..C).forEach { to ->
        var costZero = 0
        var costOne = 0
        var costTwo = 0
        (0 until N).forEach { x ->
            (0 until N).forEach { y ->
                val sum = (x + 1) + (y + 1)
                val from = c[x][y]
                val cost = D[from -1][to - 1]
                if (sum % 3 == 0) {
                    costZero += cost
                } else if (sum % 3 == 1) {
                    costOne += cost
                } else {
                    costTwo += cost
                }
            }
        }
        zeroMod3Scores[to - 1] = costZero
        oneMod3Scores[to - 1] = costOne
        twoMod3Scores[to - 1] = costTwo
    }

    var minCost = Int.MAX_VALUE
    // O(C^3)
    (0 until C).forEach {i ->
        (0 until C).forEach {j ->
            (0 until C).forEach {k ->
                if (i != j && j != k && k != i) {
                    minCost = Math.min(minCost,
                            zeroMod3Scores[i] + oneMod3Scores[j] + twoMod3Scores[k]
                    )
                }
            }
        }
    }
    println(minCost)
}