Problem statement

Of the string set $ S $ that meets the following conditions, configure $ 1 $ with the largest number of elements.

* The length of the string contained in $ S $ is $ 1 $ or more and $ N $ or less.
* The lengths of the strings contained in $ S $ are different.
* The string contained in $ S $ consists only of the characters `0` and` 1`.
* For any $ 2 $ string contained in $ S $, one is not a substring of the other.



However, the substring of the string $ x $ is the string obtained by removing more than $ 0 $ characters from the beginning and end of $ x $.

Constraint

* $ 1 \ leq N \ leq 300 $
* $ N $ is an integer



* * *

input

Input is given from standard input in the following format.


$ N $


output

Output $ 1 $ of the string set $ S $ that satisfies the condition with the maximum number of elements. There may be multiple such $ S $, but any output will be correct.

Specifically, output in the following format.

Output $ K $, the number of elements of $ S $, on the $ 1 $ line.

Output all $ S $ elements from the $ 2 $ line to the $ K + 1 $ line. Print $ 1 $ of $ S $ elements on each line.


$ K $
$ S_1 $
$ S_2 $
$ \ vdots $
$ S_K $


* * *

Input example 1


2


Output example 1


2
0
11


The string set $ \\ {0, 11 \\} $ is the set with the largest number of elements that meets the conditions.

* * *

Input example 2


1


Output example 2


1
0






Example

Input

2


Output

2
0
11
#include <bits/stdc++.h>
using namespace std;
#define MOD 1000000007
//#define MOD 998244353
#define INF 1145141919810893364
//#define INF 810114514
//#define PI 3.141592653589
typedef pair<int,int> PP;
typedef long long ll;
#define int ll
#define setdouble setprecision
#define REP(i,n) for(int i=0;i<(n);++i)
#define OREP(i,n) for(int i=1;i<=(n);++i)
#define RREP(i,n) for(int i=(n)-1;i>=0;--i)
#define GOODBYE do { cout << "0" << endl; return 0; } while (false)
#define MM <<" "<<
#define Endl endl

signed main(void){
    int N;
    cin >> N;
    if(N==1){
        cout << 1 << endl;
        cout << 0 << endl;
        return 0;
    }
    if(N==2){
        cout << 2 << endl;
        cout << 0 << endl;
        cout << 11 << endl;
        return 0;
    }
    cout << N-1 << endl;
    REP(i,N-1){
        REP(j,i+2){
            if(j==0 || j==i+1){
                cout << 1;
            }else{
                cout << 0;
            }
        }cout << endl;
    }
    
    return 0;
}


