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 C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdint.h> // uint64_t

#define max(a,b) ((a) > (b) ? (a) : (b))
#define min(a,b) ((a) > (b) ? (b) : (a))

int get_int2(int *a1, int *a2) {
  scanf("%d %d", a1, a2);
  return 0;
}

// <arr[0]> <arr[1]> .. <arr[size-1]>
int fget_array(int *arr, int size) {
    int i;
    for(i = 0; i < size; i++) {
        scanf("%d", &arr[i]);
    }
    return 0;
}

#define NUM_MAX 500
// max: 250000 * 1000
#define INF (1e+9)

int calc(int mat[][NUM_MAX+1], int *cs, int ds[][NUM_MAX+1], int size) {
    int i, j;
    int disorder = 0;
    for(i = 1; i <= size; i++) {
        for(j = 1; j <= size; j++) {
            int after = cs[(i+j)%3];
            int before = mat[i][j];
            disorder += ds[before][after];
        }
    }
#ifdef DEBUG
    printf("%d, %d, %d -> %d\n", cs[0], cs[1], cs[2], disorder);
#endif
    return disorder;
}

int main(void) {
    int num, colors;
    get_int2(&num, &colors);
    // 1-indexed
    static int ds[NUM_MAX+1][NUM_MAX+1];
    static int mat[NUM_MAX+1][NUM_MAX+1];
    int i;
    for(i = 1; i <= colors; i++) {
        fget_array(&ds[i][1], colors);
    }

    for(i = 1; i <= num; i++) {
        fget_array(&mat[i][1], num);
    }

    int c0, c1, c2; // c_(mod)
    int ans = INF;
    for(c0 = 1; c0 <= colors; c0++) {
        for(c1 = 1; c1 <= colors; c1++) {
            if(c0 == c1) continue;
            for(c2 = 1; c2 <= colors; c2++) {
                if(c1 == c2) continue;
                if(c0 == c2) continue;
                int c[3] = {c0, c1, c2};
                int res = calc(mat, c, ds, num);
                ans = min(ans, res);
            }
        }
    }
    printf("%d\n", ans);
    return 0;
}