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:
/*
Problem : AtCoder Grand Contest 030 - C
Algorithm : math
Status : AC
*/
#include <bits/stdc++.h>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <iostream>
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;

int k;

int main(){
    cin >> k;
    if (k == 1) {
        printf("1\n1\n");
        return 0;
    }
    int n = (k + 3) / 4;
    n *= 2;
    int a = 2 * n - k;
    printf("%d\n", n);
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            int k = (i + j) % n;
            printf("%d ", (i % 2 && k >= a) * (n - a) + k + 1);
        }
        printf("\n");
    }
	return 0;
}