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>
#include <unistd.h>
 
using namespace std;
 
typedef long long ll;
typedef pair<ll,ll> P;
 
long long int INF = 3e18;
const ll fact_table = 1000008;
double Pi = 3.1415926535897932384626;
 
priority_queue <ll> pql;
priority_queue <P> pqp;
//big priority queue
priority_queue <ll,vector<ll>,greater<ll> > pqls;
priority_queue <P,vector<P>,greater<P> > pqps;
//small priority queue
//top pop
 
ll dx[8]={1,0,-1,0,1,1,-1,-1};
ll dy[8]={0,1,0,-1,1,-1,-1,1};
char dir[] = "DRUL";
//ll bit[500005];
//↓,→,↑,←
 
 
#define p(x) cout<<x<<"\n";
#define el cout<<"\n";
#define pe(x) cout<<(x)<<" ";
#define ps(x) cout<<fixed<<setprecision(8)<<x<<"\n";
#define pu(x) cout<<(x);
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define CLEAR(a) a = decltype(a)(); 
 
 
ll mod = 998244353;
//ll mod = 1000000007;
 
ll rui(ll number1,ll number2){
 
    if(number2 == 0){
        return 1;
    }else{
        ll number3 = rui(number1,number2 / 2);
        number3 *= number3;
        number3 %= mod;
        if(number2%2==1){
            number3 *= number1;
            number3 %= mod;
        }
        return number3;
    }
}
ll mod1 = 2000000011;
ll mod2 = 2000000033;


ll rui1(ll number1,ll number2){
 
    if(number2 == 0){
        return 1;
    }else{
        ll number3 = rui1(number1,number2 / 2);
        number3 *= number3;
        number3 %= mod1;
        if(number2%2==1){
            number3 *= number1;
            number3 %= mod1;
        }
        return number3;
    }
}

ll rui2(ll number1,ll number2){
 
    if(number2 == 0){
        return 1;
    }else{
        ll number3 = rui2(number1,number2 / 2);
        number3 *= number3;
        number3 %= mod2;
        if(number2%2==1){
            number3 *= number1;
            number3 %= mod2;
        }
        return number3;
    }
}
ll gcd(ll number1,ll number2){
 
    if(number1 > number2){
        swap(number1,number2);
    }
    if(number1 == 0 || number1 == number2){
        return number2;
    }else{
        return gcd(number2 % number1,number1);
    }
}
void YES(bool condition){
 
    if(condition){
        p("YES");
    }else{
        p("NO");
    }
    return;
}
void Yes(bool condition){
 
    if(condition){
        p("Yes");
    }else{
        p("No");
    }
    return;
}
 
/*
ll fact[fact_table + 5],rfact[fact_table + 5];
 
void c3_init(){
    fact[0] = rfact[0] = 1;
    for(ll i=1; i<=fact_table; i++){
        fact[i] = (fact[i-1]*i) % mod;
    }
    rfact[fact_table] = rui(fact[fact_table],mod - 2);
    for(ll i=fact_table; i>=1; i--){
       rfact[i-1] = rfact[i] * i;
       rfact[i-1] %= mod;
    }
    return;}
ll c3(ll n,ll r){
    return (((fact[n] * rfact[r]) % mod ) * rfact[n-r]) % mod;}
*/

struct edge {ll to,cost,idx; } ;

void fast(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    return;
}
 
ll n,m,k,num,a,b,c,d,e,g,h,w,i,j,q,r,l;
ll x[50005];
ll dp[500][1020];

void solve(){
    cin >> n;
    if(n == 1){
        p(1);
        p(0);
        return;
    }else if(n == 2){
        p(2);
        p(0);
        p(11);
        return;
    }else{
        p(n-1);
        for(ll i=0;i<n-1;i++){
            pu(0);
            for(ll j=0;j<i;j++){
                pu(1);
            }
            pu(0);
            el;
        }
        return;
    }
    return;
}

int main(){
    // fast();
    q = 1;
    while(q--){
        solve();
    }
    return 0;
}










