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<cstring>
#include<cstdio>
#include<iostream>
using namespace std;
int main() {
	int k;
	cin >> k;
	if(k<=2) {
		cout << k << endl;
		for(int i=1; i<=k; i++) {
			for(int j=1; j<=k; j++) {
				if(j==k) printf("%d",j);
				else printf("%d ",j);
			}
			cout << endl;
		}
		return 0;
	}
	int n=k+1;
	n>>=1;
	if(n%2==1) n++;
	cout << n << endl;
	int cnt1=1,cnt2=2;
	for(int i=1; i<=n; i++) {
		for(int j=1; j<=n; j++) {
			int tar;
			if(i&1) {
				tar=(cnt1+j)%n;
				if(tar>=k) tar-=n;
			} else {
				tar=n+(cnt2+j)%n;
				if(tar>=k) tar-=n;

			}
			if(j==n) printf("%d",1+tar);
			else printf("%d ",1+tar);
		}
		if(i&1) cnt1+=2;
		else cnt2+=2;
		printf("\n");
	}
}