#include <bits/stdc++.h>
using namespace std;
const int N = 505;
int c[N][N],n;
int main() {
	scanf("%d",&n);
	if(n==1) printf("Yes\n1\n");
	else {
		if(n&1) printf("No\n");
		else {
			printf("Yes\n");
			for(int i=1;i<n;i++) c[i][1]=c[n-1][i+1]=i;
			for(int i=n-2;i>=2;i--) {
				for(int j=2;j<=i;j++) c[i][j]=c[i+1][j-1];
			}
			for(int i=1;i<n;i++) c[n][i]=c[i][i];
			for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++)
				if(i<j) c[i][j]=c[j][i]+n-1;
			for(int i=1;i<=n;i++) c[i][i]=2*n-1;
			for(int i=1;i<=n;i++) 
				for(int j=1;j<=n;j++) printf("%d%c",c[i][j],j==n?'\n':' '); 
		}
	}
	return 0;
}


### Prompt

Please provide a cpp coded solution to the problem described below:

J: Horizontal-Vertical Permutation

Problem Statement

You are given a positive integer N. Your task is to determine if there exists a square matrix A whose dimension is N that satisfies the following conditions and provide an example of such matrices if it exists. A_{i, j} denotes the element of matrix A at the i-th row and j-th column.

* For all i, j (1 \leq i, j \leq N), A_{i, j} is an integer that satisfies 1 \leq A_{i, j} \leq 2N - 1.
* For all k = 1, 2, ..., N, a set consists of 2N - 1 elements from the k-th row or k-th column is \\{1, 2, ..., 2N - 1\\}.



If there are more than one possible matrices, output any of them.

Input


N

Input consists of one line, which contains the integer N that is the size of a square matrix to construct.

Constraint

* N is an integer that satisfies 1 \leq N \leq 500.



Output

Output `No` in a single line if such a square matrix does not exist.

If such a square matrix A exists, output `Yes` on the first line and A after that. More specifically, follow the following format.


Yes
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
:
A_{N, 1} A_{N, 2} ... A_{N, N}


Sample Input 1


4

Output for Sample Input 1


Yes
2 6 3 7
4 5 2 1
1 7 5 6
5 3 4 2


Sample Input 2


3

Output for Sample Input 2


No





Example

Input

4


Output

Yes
2 6 3 7
4 5 2 1
1 7 5 6
5 3 4 2

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 505;
int c[N][N],n;
int main() {
	scanf("%d",&n);
	if(n==1) printf("Yes\n1\n");
	else {
		if(n&1) printf("No\n");
		else {
			printf("Yes\n");
			for(int i=1;i<n;i++) c[i][1]=c[n-1][i+1]=i;
			for(int i=n-2;i>=2;i--) {
				for(int j=2;j<=i;j++) c[i][j]=c[i+1][j-1];
			}
			for(int i=1;i<n;i++) c[n][i]=c[i][i];
			for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++)
				if(i<j) c[i][j]=c[j][i]+n-1;
			for(int i=1;i<=n;i++) c[i][i]=2*n-1;
			for(int i=1;i<=n;i++) 
				for(int j=1;j<=n;j++) printf("%d%c",c[i][j],j==n?'\n':' '); 
		}
	}
	return 0;
}


```