#include <iostream>

using namespace std;

int n,c,ans = 1000000000;
int d[1001][1001];
int cc[501][501];
int co[3][31];


int main()
{
  

  cin >> n >> c;

  for(int i = 1;i <= c;i++){
    for(int j = 1;j <= c;j++){
      cin >> d[i][j];
    }
  }

  for(int i = 1;i <= n;i++){
    for(int j = 1;j <= n;j++){
      cin >> cc[i][j];
      co[(i+j)%3][cc[i][j]]++;
    }
  }

  for(int i = 1;i <= c;i++){
    for(int j = 1;j <= c;j++){
      for(int k = 1;k <= c;k++){
	if(i == j || j == k || k == i) continue;
	int ai = 0,aj = 0,ak = 0;
	for(int a = 1;a <= c;a++){
	  ai += co[0][a] * d[a][i];
	  aj += co[1][a] * d[a][j];
	  ak += co[2][a] * d[a][k];
	}
	if(ans > ai + aj + ak)
	  ans = ai + aj + ak;
      }
    }
  }

  cout << ans << endl;

  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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.

Examples

Input

2 3
0 1 1
1 0 1
1 4 0
1 2
3 3


Output

3


Input

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


Output

428

### Response

```cpp
#include <iostream>

using namespace std;

int n,c,ans = 1000000000;
int d[1001][1001];
int cc[501][501];
int co[3][31];


int main()
{
  

  cin >> n >> c;

  for(int i = 1;i <= c;i++){
    for(int j = 1;j <= c;j++){
      cin >> d[i][j];
    }
  }

  for(int i = 1;i <= n;i++){
    for(int j = 1;j <= n;j++){
      cin >> cc[i][j];
      co[(i+j)%3][cc[i][j]]++;
    }
  }

  for(int i = 1;i <= c;i++){
    for(int j = 1;j <= c;j++){
      for(int k = 1;k <= c;k++){
	if(i == j || j == k || k == i) continue;
	int ai = 0,aj = 0,ak = 0;
	for(int a = 1;a <= c;a++){
	  ai += co[0][a] * d[a][i];
	  aj += co[1][a] * d[a][j];
	  ak += co[2][a] * d[a][k];
	}
	if(ans > ai + aj + ak)
	  ans = ai + aj + ak;
      }
    }
  }

  cout << ans << endl;

  return 0;
}

```