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>

const int MAXN = 1010;
int A[2][MAXN];
int K, n;
void shift(int * A) {
	memmove(A, A + 1, n << 2);
	A[n] = A[0];
}
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> K; n = (K + 3 >> 2) << 1;
	if (K == 1) return std::cout << "1\n1\n", 0;
	auto red = [] (int x) { return x > K ? x - n : x; };
	int idx = 0;
	for (int k = 0; k < 2; ++k)
		for (int j = 1; j <= n; ++j)
			A[k][j] = ++idx;
	shift(A[1]);
	std::cout << n << '\n';
	for (int i = 1; i <= n / 2; ++i) {
		for (int k = 0; k < 2; ++k)
			for (int j = 1; j <= n; ++j)
				std::cout << red(A[k][j]) << (" \n" [j == n]);
		shift(A[0]); shift(A[0]);
		shift(A[1]); shift(A[1]);
	}
	return 0;
}