For an n \times n grid, let (r, c) denote the square at the (r+1)-th row from the top and the (c+1)-th column from the left. A good coloring of this grid using K colors is a coloring that satisfies the following:

* Each square is painted in one of the K colors.
* Each of the K colors is used for some squares.
* Let us number the K colors 1, 2, ..., K. For any colors i and j (1 \leq i \leq K, 1 \leq j \leq K), every square in Color i has the same number of adjacent squares in Color j. Here, the squares adjacent to square (r, c) are ((r-1)\; mod\; n, c), ((r+1)\; mod\; n, c), (r, (c-1)\; mod\; n) and (r, (c+1)\; mod\; n) (if the same square appears multiple times among these four, the square is counted that number of times).



Given K, choose n between 1 and 500 (inclusive) freely and construct a good coloring of an n \times n grid using K colors. It can be proved that this is always possible under the constraints of this problem,

Constraints

* 1 \leq K \leq 1000

Input

Input is given from Standard Input in the following format:


K


Output

Output should be in the following format:


n
c_{0,0} c_{0,1} ... c_{0,n-1}
c_{1,0} c_{1,1} ... c_{1,n-1}
:
c_{n-1,0} c_{n-1,1} ... c_{n-1,n-1}


n should represent the size of the grid, and 1 \leq n \leq 500 must hold. c_{r,c} should be an integer such that 1 \leq c_{r,c} \leq K and represent the color for the square (r, c).

Examples

Input

2


Output

3
1 1 1
1 1 1
2 2 2


Input

9


Output

3
1 2 3
4 5 6
7 8 9
#include <cstdio>
using namespace std;
int k;
namespace jzp1 {
	inline void sol() {
		printf("%d\n", k);
		for (int i = 1; i <= k; ++i) {
			for (int j = 1, st = i; j <= k; ++j, st = st % k + 1)
				printf("%d ", st);
			puts("");
		}
	}
}
namespace jzp2 {
	int mp[1111];
	inline void sol() {
		int n = (k + 3) / 4 * 2;
		printf("%d\n", n);
		for (int i = 1; i <= k; ++i)	mp[i] = i;
		for (int i = k + 1; i <= n + n; ++i)	mp[i] = (i - n) % n + 1;
		for (int i = 1; i <= n; ++i) {
			for (int j = 1, st = i, lst = i - 1; j <= n; ++j, lst = st, st = st % n + 1) {
				if (i & 1)	printf("%d ", mp[st]);
				else	printf("%d ", mp[lst + n]);
			}
			puts("");
		}
	}
}
int main() {
	scanf("%d", &k);
	if (k <= 500) jzp1::sol();
	else
		jzp2::sol();
	return 0;
}
