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
#pragma GCC optimize ("O3")
#include <iostream>
#include <iomanip>
#include <istream>
#include <ostream>
#include <sstream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <bitset>
#include <utility>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <ctime>
#include <cctype>
#include <cstdlib>
#define IINF 10e8
#define INF 1<<30
#define MOD 1000000007
#define mod 998244353
#define REP(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define REPE(i, a, n) for (ll i = a; i <= (ll)(n); i++)
#define Endl endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define mmax(x,y)(x>y?x:y)
#define mmin(x,y)(x<y?x:y)
#define chmax(x,y) x=mmax(x,y)
#define chmin(x,y) x=mmin(x,y)
#define all(x) (x).begin(),(x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
using namespace std;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int>Pin;
typedef pair<ll,ll>Pll;
template<class T> using V=vector<T>;
long long GCD(long long a, long long b) {return b?GCD(b,a%b):a;}
long long LCM(long long a, long long b) {return a/GCD(a,b)*b;}
int dx[4]={-1,0,1,0};
int dy[4]={0,-1,0,1};
int ddx[8]={-1,0,1,0,1,1,-1,-1};
int ddy[8]={0,-1,0,1,1,-1,1,-1};
ll cmp(pair<ll,ll>a,pair<ll,ll> b){
        if(a.se!=b.se)
        return a.se<b.se;
        else
        return a.fi<b.fi;
}
//----------------------------------------------------------------------

//----------------------------------------------------------------------
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    //-------------------------------
    int n;cin>>n;
    if(n==1){
        cout<<1<<endl;
        cout<<0<<endl;
        return 0;
    }
    else if(n==2){
        cout<<2<<endl;
        cout<<0<<endl;
        cout<<11<<Endl;
        return 0;
    }
    int zero=1;int pos=0;
    int cnt=0;
    V<string>ans;
    int len=n;
    /*for(len=n;len>0;len--){
        string s="";
        for(int i=0;i<len;i++){
            s+='1';
        }
        if(len==n){
            ans.pb(s);
            cnt++;
            continue;
        }
        int nowpos=len-1-pos-(zero-1);
        if(nowpos>=0){
            for(int j=nowpos;j<nowpos+zero;j++)
                s[j]='0';
            pos++;
        }
        else{
            zero++;
            pos=0;
            nowpos=len-1-pos-(zero-1);
            if(nowpos>=0){
                for(int j=nowpos;j<nowpos+zero;j++)
                    s[j]='0';
                pos++;
            }
            else{
                break;
            }
        }
        ans.pb(s);
        cnt++;
    }*/
    zero=1;pos=0;
    for(len=len;len>0;len--){
        string s="";
        for(int i=0;i<len;i++){
            s+='1';
        }
        s[0]='0';
        s[len-1]='0';
        if(len==1)break;
        ans.pb(s);
        cnt++;
    }
    cout<<cnt<<endl;
    for(int i=0;i<ans.size();i++){
        cout<<ans[i]<<endl;
    }
    //-------------------------------
    return 0;
}
//----------------------------------------------------------------------

































