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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <inttypes.h>
#include <stdint.h>
#include <string.h>
#include <wchar.h>

#define MIN(a, b) (a < b ? a : b)
#define MAX(a, b) (a > b ? a : b)

static uint32_t D[30][30];
static uint32_t c[500][500];

static uint32_t calcStrangeScore(const uint32_t N, const uint32_t C, const uint32_t zero, const uint32_t one, const uint32_t two) {
	uint32_t strangeScore = 0;
	
	for (uint32_t row = 0; row < N; row++) {
		for (uint32_t col = 0; col < N; col++) {
			const uint32_t remainder = (row + 1 + col + 1) % 3;
			if (remainder == 0) {
				if (c[row][col] != zero) {
					strangeScore += D[c[row][col] - 1][zero - 1];
				}
			}
			else if (remainder == 1) {
				if (c[row][col] != one) {
					strangeScore += D[c[row][col] - 1][one - 1];
				}

			}
			else {
				if (c[row][col] != two) {
					strangeScore += D[c[row][col] - 1][two - 1];
				}

			}
		}
	}

	return strangeScore;
}

int main(void) {
	uint32_t N, C;

	scanf("%"PRIu32" %"PRIu32"\n", &N, &C);

	for (uint32_t dRow = 0; dRow < C; dRow++) {
		for (uint32_t dCol = 0; dCol < C - 1; dCol++) {
			scanf("%"PRIu32" ", &(D[dRow][dCol]));
		}
		scanf("%"PRIu32"\n", &(D[dRow][C - 1]));
	}

	for (uint32_t cRow = 0; cRow < N; cRow++) {
		for (uint32_t cCol = 0; cCol < N - 1; cCol++) {
			scanf("%"PRIu32" ", &(c[cRow][cCol]));
		}
		scanf(cRow == N - 1 ? "%"PRIu32 : "%"PRIu32"\n", &(c[cRow][N - 1]));
	}

	uint32_t min = UINT32_MAX;

	for (uint32_t zero = 1; zero <= C; zero++) {
		for (uint32_t one = 1; one <= C; one++) {
			if (one != zero) {
				for (uint32_t two = 1; two <= C; two++) {
					if (two != zero && two != one) {
						const uint32_t strangeScore = calcStrangeScore(N, C, zero, one, two);
						if (strangeScore < min) {
							min = strangeScore;
						}
					}
				}
			}
		}
	}

	printf("%"PRIu32, min);

	return 0;
}