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;

const int N = 505;

int k;
int res[N][N];
int id[2*N], cur = 1;

int get(int i){
    if(id[i] == 0) return id[i] = cur++;
    else return id[i];
}

int main(){
    cin >> k;
    if(k == 1) return !printf("1\n1\n");
    int n = (k + 1) / 2;
    if(n % 2 != 0) n++;
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            if(i % 2 == 1) res[i][j] = (i + j) % n;
            else res[i][j] = (i + j) % n + n;
        }
    }
    int cnt = 0;
    while(k < 2 * n){
        for(int i = 0; i < n; i++) res[i][(cnt-i + n) % n] = cnt;
        k++;
        cnt++;
    }
    cout << n << endl;
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++) cout << get(res[i][j]) << "\t";
        cout << endl;
    }
}