John Doe started thinking about graphs. After some thought he decided that he wants to paint an undirected graph, containing exactly k cycles of length 3.

A cycle of length 3 is an unordered group of three distinct graph vertices a, b and c, such that each pair of them is connected by a graph edge.

John has been painting for long, but he has not been a success. Help him find such graph. Note that the number of vertices there shouldn't exceed 100, or else John will have problems painting it.

Input

A single line contains an integer k (1 ≤ k ≤ 10⁵) — the number of cycles of length 3 in the required graph.

Output

In the first line print integer n (3 ≤ n ≤ 100) — the number of vertices in the found graph. In each of next n lines print n characters "0" and "1": the i-th character of the j-th line should equal "0", if vertices i and j do not have an edge between them, otherwise it should equal "1". Note that as the required graph is undirected, the i

-th character of the j-th line must equal the j-th character of the i-th line. The graph shouldn't contain self-loops, so the i-th character of the i-th line must equal "0" for all i.

Sample test(s) Input

1

Output

3
011
101
110

Input

10

Output

5
01111
10111
11011
11101
11110

題目大意：

已知圖中的三條邊可以形成一個三角形。現給定k個三角形(k<10^5)，求圖的鄰接矩陣。節點數n<=100;

解題思路：

首先用排列組合的方法確定最少的完全圖。然後計算，每新增一個點，將與完全圖中的兩個點形成一個新的三角形。

通過這種方法，貪心把圖求出來。

#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;

int main()
{
    int f[111][111];
    memset(f,0,sizeof(f));
    f[0][0]=f[1][0]=f[1][1]=f[2][2]=f[2][0]=1;
    f[2][1]=2;
    for( int i=3;i<=100;i++ )
    {
        for( int j=0;j<=3;j++ )
            f[i][j]=f[i-1][j-1]+f[i-1][j];
    }
    int n;
    while( scanf("%d",&n)!=EOF )
    {
        int i,j,ans,m[111][111];
        memset( m,0,sizeof(m) );
        for( i=2;n>=f[i+1][3];i++ ) ;
        n-=f[i][3];
        //printf("n=%d %d\n",n,i);
        ans=i;
        for( i=1;i<=ans;i++ ) for( j=1;j<=ans;j++ )
            m[i][j]=i==j?0:1;
        while( n )
        {
            for( i=2;n>=f[i+1][2];i++ ) ;
            n-=f[i][2]; ans++;
            for( j=1;j<=i;j++ )
                m[ans][j]=m[j][ans]=1;
        }
        printf("%d\n",ans);
        for( i=1;i<=ans;i++ )
        {
            for( j=1;j<=ans;j++ )
                printf("%d",m[i][j]);
            printf("\n");
        }
    }
    return 0;
}