Problem:
Score : 1100 points

Problem Statement

You are given an integer N.

Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.

1 \leq a_{i,j} \leq 10^{15}

a_{i,j} are pairwise distinct integers.

There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.

Constraints

2 \leq N \leq 500

Input

Input is given from Standard Input in the following format:

N

Output

Print your solution in the following format:

a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Sample Input 1

2

Sample Output 1

4 7
23 10

For any two elements x and y that are vertically or horizontally adjacent, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always 3.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
int n,a[505][505],p[30005],vis[30005];
ll gcd(ll x,ll y){
	if (!y)return x;
	return gcd(y,x%y);
}
ll lcm(ll x,ll y){
	if ((!x)||(!y))return x+y;
	return x*y/gcd(x,y);
}
int main(){
	scanf("%d",&n);
	for(int i=2;i<=30000;i++){
		if (!vis[i])p[++p[0]]=i;
		for(int j=1;(j<=p[0])&&(i*p[j]<=30000);j++){
			vis[i*p[j]]=1;
			if (i%p[j]==0)break;
		}
	}
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			if ((i+j)%2)a[i][j]=p[i-j+n/2*2+1]*p[(i+j)/2];
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			if ((i+j)%2==0)a[i][j]=lcm(lcm(a[i-1][j],a[i][j-1]),lcm(a[i+1][j],a[i][j+1]))+1;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++)printf("%lld ",a[i][j]);
		printf("\n");
	}
}