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 <bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef long double llf;
typedef pair<int, int> pi;
const int MAXN = 605;
const int mod = 1e9 + 7;
#define sz(v) ((int)v.size())

int n, a[MAXN][MAXN];

int main(){
	cin >> n;
	if(n <= 500){
		cout << n << endl;
		for(int i=0; i<n; i++){
			for(int j=0; j<n; j++){
				printf("%d ", (i + j) % n + 1);
			}
			puts("");
		}
		return 0;
	}
	else{
		int k = (n + 1) / 2;
		if(k % 2 == 1) k++;
		for(int i=0; i<k; i++){
			for(int j=0; j<k; j++){
				a[i][j] = (i + j) % k + 1;
				if(j % 2 == 1) a[i][j] += k;
				if(a[i][j] > n) a[i][j] -= k;
			}
		}
		cout << k << endl;
		for(int i=0; i<k; i++){
			for(int j=0; j<k; j++) printf("%d ", a[i][j]);
			puts("");
		}
	}
}