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>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n;
bool prime[50000005];
int ans[501][501];
vector<ll> vec;

int main(void){
	scanf("%d",&n);
	memset(prime,true,sizeof(prime));
	for(ll i=2;i<50000005;i++){
		if(prime[i]){
			vec.push_back(i);
			for(ll j=2LL*i;j<50000005;j+=i){
				prime[j]=false;
			}
		}
	}
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			printf("%lld%c",vec[i*n+j]*vec[n*n]+vec[n*n+1],j+1==n?'\n':' ');
		}
	}
	return 0;
}