Problem: 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

Solution (in CPP):

Solution:
#include <cstdio>

const int N = 500;

int a[N][N];

int main() {
	int K; scanf("%d", &K);
	if( K <= N ) {
		printf("%d\n", K);
		for(int i=1;i<=K;i++) {
			for(int j=1;j<=K;j++)
				printf("%d%c", i, (j == K ? '\n' : ' '));
		}
	}
	else {
		printf("%d\n", N);
		int cnt = 0;
		for(int i=0;i<2*N-K;i++) {
			int x = 0, y = i, b = (++cnt);
			do {
				a[x][y] = b;
				x = (x + 1 == N ? 0 : x + 1);
				y = (y + 1 == N ? 0 : y + 1);
			}while( x );
		}
		for(int i=2*N-K;i<N;i++) {
			int x = 0, y = i, b[2] = {};
			b[0] = (++cnt), b[1] = (++cnt);
			do {
				a[x][y] = b[x & 1];
				x = (x + 1 == N ? 0 : x + 1);
				y = (y + 1 == N ? 0 : y + 1);
			}while( x );
		}
		for(int i=0;i<N;i++) {
			for(int j=0;j<N;j++)
				printf("%d%c", a[i][j], (j + 1 == N ? '\n' : ' '));
		}
	}
}