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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 500, dir[4][2] = {0, 1, 1, 0, 0, -1, -1, 0};
int n, m, mn;
ll s[maxn * maxn + 10], a[maxn + 10][maxn + 10];

bool ispr(int x) {
	for (int i = 2; i * i <= x; ++i)
		if (x % i == 0) return 0;
	return 1;
}

int main() {
	scanf("%d", &n);
	if (n == 2) {
		printf("%d %d\n%d %d\n", 4, 7, 23, 10);
		return 0;
	}
	for (int i = 1, p = 1; i <= n * 2; ++i) {
		for (++p; !ispr(p); ++p);
		s[i] = p;
	}
	mn = (1 - n) & 1 ? (2 - n) / 2 : (1 - n) / 2;
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= n; ++j)
			if ((i + j) % 2 == 0) {
				int x = (i + j) / 2, y = (i - j) / 2 - mn + 1;
				a[i][j] = s[x] * s[y + n];
			}
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= n; ++j) {
			if ((i + j) % 2) {
				ll g = 1;
				for (int k = 0; k < 4; ++k) {
					int dx = i + dir[k][0], dy = j + dir[k][1];
					if (dx >= 1 && dx <= n && dy >= 1 && dy <= n)
						g = g / __gcd(g, a[dx][dy]) * a[dx][dy];
				}
				a[i][j] = g + 1;
			}
			printf("%lld%c", a[i][j],  " \n"[j == n]);
		}
}