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 <bits/stdc++.h>
#define ll long long
#define rep2(i,a,b) for(ll i=a;i<=b;++i)
#define rep(i,n) for(ll i=0;i<n;i++)
#define rep3(i,a,b) for(ll i=a;i>=b;i--)
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pq priority_queue
#define pb push_back
#define eb emplace_back
#define vec vector<int>
#define vecll vector<ll>
#define vecpii vector<pii>
#define vec2(a,b) vector<vec>(a,vec(b))
#define vec2ll(a,b) vector<vec>(a,vecll(b))
#define vec3(a,b,c) vector<vector<vec>>(a,vec2(b,c))
#define vec3ll(a,b,c) vector<vector<vecll>>(a,vec2ll(b,c))
#define fi first
#define se second
#define endl "\n"
#define all(c) begin(c),end(c)
#define ios ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define lb(c,x) distance(c.begin(),lower_bound(all(c),x))
#define ub(c,x) distance(c.begin(),upper_bound(all(c),x))
#define MM " "
using namespace std;
int in() {int x;cin>>x;return x;}
ll lin() {ll x;cin>>x;return x;}
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
template<class T> inline void print(pair<T,T> p){cout<<"("<<p.first<<","<<p.second<<") ";}
template<class T> inline void print(vector<pair<T,T>> v){for(auto e:v)print(e); cout<<endl;}
template<class T> inline void print(T v){for(auto e:v)cout<<e<<" ";cout<<endl;}


int s[4][4]={
    {6,3,1,4},
    {2,0,2,0},
    {1,3,6,4},
    {5,0,5,0}
};
int check(int x,int y){
    return s[x%4][y%4];
}

int main(){
    int n=in();
    if(n==1){
        cout<<1<<endl<<0<<endl;
        return 0;
    }
    else if(n==2) {
        cout<<2<<endl<<"00"<<endl<<1<<endl;
        return 0;
    }
    cout<<n-1<<endl;
    rep2(i,2,n){
        cout<<1;
        rep(j,i-2)cout<<0;
        cout<<1<<endl;
    }
    
}
