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:
#pragma GCC optimize(3)
#include<bits/stdc++.h>
#define MAXN 100005
#define INF 1000000000
#define MOD 1000000007
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int n,k;
int ans[505][505];
int main()
{
    scanf("%d",&k);
    if(k==1)
    {
        printf("1\n1\n");
        return 0;
    }
    n=(k+3)/4*2;
    printf("%d\n",n);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            ans[i][j]=(i+j-1)%n+1;
            if(i&1) ans[i][j]+=n;
        }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            if(ans[i][j]>k) ans[i][j]-=n;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
            printf("%d ",ans[i][j]);
        puts("");
    }
    return 0;
}