#include"bits/stdc++.h"
using namespace std;
#define LL long long
#define PB push_back
#define MP make_pair

int N, C, D[30][30];
int c[3][30] ={};
int main(){
	cin >> N >> C;
	for(int i=0; i<C; i++){
		for(int j=0; j<C; j++){
			cin >> D[i][j];
		}
	}
	for(int i=0; i<N; i++){
		for(int j=0; j<N; j++){
			int tmp;
			cin >> tmp;
			c[(i+j)%3][tmp-1]++;
		}
	}

	int ans = INT_MAX;
	for(int i=0; i<C; i++){
		for(int j=0; j<C; j++){
			for(int k=0; k<C; k++){
				if(i == j || i == k || j == k) continue;
				int t[3]={ i,j,k };
				int sum = 0;
				for(int l=0; l<3; l++){
					for(int n=0; n<C; n++){
						sum += D[n][t[l]] * c[l][n];
					}
				}
				ans = min(ans, sum);
			}
		}
	}
	cout << ans << endl;
}
### Prompt

Please create a solution in CPP to the following problem:

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"bits/stdc++.h"
using namespace std;
#define LL long long
#define PB push_back
#define MP make_pair

int N, C, D[30][30];
int c[3][30] ={};
int main(){
	cin >> N >> C;
	for(int i=0; i<C; i++){
		for(int j=0; j<C; j++){
			cin >> D[i][j];
		}
	}
	for(int i=0; i<N; i++){
		for(int j=0; j<N; j++){
			int tmp;
			cin >> tmp;
			c[(i+j)%3][tmp-1]++;
		}
	}

	int ans = INT_MAX;
	for(int i=0; i<C; i++){
		for(int j=0; j<C; j++){
			for(int k=0; k<C; k++){
				if(i == j || i == k || j == k) continue;
				int t[3]={ i,j,k };
				int sum = 0;
				for(int l=0; l<3; l++){
					for(int n=0; n<C; n++){
						sum += D[n][t[l]] * c[l][n];
					}
				}
				ans = min(ans, sum);
			}
		}
	}
	cout << ans << endl;
}
```