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;
typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<ll, P>IP;
typedef vector<ll> V;
typedef vector<V> V2;
typedef vector<vector<P> > G;
void g_dir(G &graph, ll a, ll b, ll w = 1){graph[a].push_back(P(b, w));}
void g_undir(G &graph, ll a, ll b, ll w = 1){g_dir(graph, a, b, w);g_dir(graph, b, a, w);}
#define rep(i, n) for(ll (i) = 0; (i) < (n); (i)++)
#define rep1(i, n) for(ll (i) = 1; (i) <= (n); (i)++)
#define rrep(i, n) for(ll (i) = (n) - 1; (i) >= 0; (i)--)
#define rrep1(i, n) for(ll (i) = (n); (i) >= 1; (i)--)
template<class T> void chmax(T &a, const T &b){if(a < b){a = b;}}
template<class T> void chmin(T &a, const T &b){if(a > b){a = b;}}
const ll INF = 1145141919;
const ll MOD = 1000000007;
const ll NUM = 101010;

int main(){

    ll N;
    cin >> N;
    if(N == 1){
        cout << 1 << endl;
        cout << 0 << endl;
    }
    else if(N == 2){
        cout << 2 << endl;
        cout << 0 << endl;
        cout << 11 << endl;
    }
    else{
        cout << N - 1 << endl;
        rep(i, N - 1){
            cout << 1;
            rep(j, i)cout << 0;
            cout << 1 << endl;
        }
        
    }

    return 0;
}
