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:
// includes {{{
#include<iostream>
#include<iomanip>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<tuple>
#include<cmath>
#include<random>
#include<cassert>
// #include<deque>
// #include<multiset>
// #include<bitset>
// #include<cstring>
// #include<bits/stdc++.h>
// }}}
using namespace std;
using ll = long long;

int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int k;
  cin >> k;
  if(k == 1) {
    cout << "1\n1\n";
    return 0;
  }
  int n = 2 * ((k + 3) / 4);
  cout << n << "\n";
  for(int i = 0; i < n; i++) {
    for(int j = 0; j < n; j++) {
      int v = i & 1 ? (i + j) % n : n + (i + j) % n;
      if(v >= k) v -= n;
      cout << v + 1 << (j == n - 1 ? "" : " ");
    }
    cout << "\n";
  }
  return 0;
}