#include <iostream>
#include <utility>
#include <vector>

using namespace std;
typedef pair<int, int> P;

int n;
int ans[505][505];

int main(void)
{
	cin >> n;
	if(n == 1){
		cout << "Yes" << endl;
		cout << 1 << endl;
		return 0;
	}
	if(n%2){
		cout << "No" << endl;
		return 0;
	}
	for(int i = 1; i <= n-1; i++){
		for(int j = 1; j <= n-1; j++){
			if(i < j) continue;
			ans[i][j] = (i+j-2)%(n-1)+1;
		}
	}
	for(int j = 1; j <= n-1; j++){
		ans[n][j] = ans[j][j];
	}
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= n; j++){
			if(i >= j) continue;
			ans[i][j] = ans[j][i] + n-1;
		}
	}
	for(int i = 1; i <= n; i++) ans[i][i] = 2*n-1;
	
	cout << "Yes" << endl;
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= n; j++){
			cout << ans[i][j];
			if(j < n) cout << " ";
		}
		cout << endl;
	}
	
	return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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 <iostream>
#include <utility>
#include <vector>

using namespace std;
typedef pair<int, int> P;

int n;
int ans[505][505];

int main(void)
{
	cin >> n;
	if(n == 1){
		cout << "Yes" << endl;
		cout << 1 << endl;
		return 0;
	}
	if(n%2){
		cout << "No" << endl;
		return 0;
	}
	for(int i = 1; i <= n-1; i++){
		for(int j = 1; j <= n-1; j++){
			if(i < j) continue;
			ans[i][j] = (i+j-2)%(n-1)+1;
		}
	}
	for(int j = 1; j <= n-1; j++){
		ans[n][j] = ans[j][j];
	}
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= n; j++){
			if(i >= j) continue;
			ans[i][j] = ans[j][i] + n-1;
		}
	}
	for(int i = 1; i <= n; i++) ans[i][i] = 2*n-1;
	
	cout << "Yes" << endl;
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= n; j++){
			cout << ans[i][j];
			if(j < n) cout << " ";
		}
		cout << endl;
	}
	
	return 0;
}

```