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<iostream>
#include<cmath>
#include<cstdlib>
#include<string>
#include<cstring>
#include<algorithm>
#include<vector>

using namespace std;

int main() {
   int K;
   cin >> K;
   int n = (K + 3) / 4 * 2;
   if (K == 1) {
      cout << "1" << endl;
      cout << "1" << endl;
      return 0;
   }
   cout << n << endl;
   for (int r = 0; r < n; r++) {
      if (r % 2 == 0) {
         for (int c = 0; c < n; c++) {
            cout << (r + c) % n + 1;
            if (c != n - 1) cout << " ";
         }
      } else {
         for (int c = 0; c < n; c++) {
            int tmp = (r + c) % n + n + 1;
            if (tmp > K) cout << tmp - n;
            else cout << tmp;
            if (c != n - 1) cout << " ";
         }
      }
      cout << endl;
   }
   return 0;
}