#include <iostream>
#include<vector>
#include<algorithm>
using namespace std;
#define rep(i,n) for(int i = 0;i <n;i++)
int main(){
	int n, c,ans = 1<<30; cin >> n >> c;
	vector<vector<int>> D(c, vector<int>(c)),C(n,vector<int>(n)),t(3,vector<int>(c+1,0));
	rep(i, c) rep(j, c) cin >> D[i][j];
	rep(i, n) rep(j, n) cin >> C[i][j], t[(i + j) % 3][C[i][j] - 1]++;
	rep(i, c) rep(j, c) if (i != j) rep(k, c) {
		if (i != k && j != k) {
			int tt = 0;
			rep(l, c) tt += t[0][l] * D[l][i];
			rep(l, c) tt += D[l][j] * t[1][l];
			rep(l, c) tt += D[l][k] * t[2][l];
			if (tt < ans) ans = tt;
		}
	}cout << ans << endl;
}
### Prompt

Generate a Cpp solution 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 <iostream>
#include<vector>
#include<algorithm>
using namespace std;
#define rep(i,n) for(int i = 0;i <n;i++)
int main(){
	int n, c,ans = 1<<30; cin >> n >> c;
	vector<vector<int>> D(c, vector<int>(c)),C(n,vector<int>(n)),t(3,vector<int>(c+1,0));
	rep(i, c) rep(j, c) cin >> D[i][j];
	rep(i, n) rep(j, n) cin >> C[i][j], t[(i + j) % 3][C[i][j] - 1]++;
	rep(i, c) rep(j, c) if (i != j) rep(k, c) {
		if (i != k && j != k) {
			int tt = 0;
			rep(l, c) tt += t[0][l] * D[l][i];
			rep(l, c) tt += D[l][j] * t[1][l];
			rep(l, c) tt += D[l][k] * t[2][l];
			if (tt < ans) ans = tt;
		}
	}cout << ans << endl;
}
```