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 <iostream>
#include <algorithm>
#include <utility>
#include <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>
#include <math.h>
#include <string.h>
#include <iomanip>
#include <numeric>
#include <cstdlib>
#include <cstdint>
#include <cmath>
#include <functional>
#include <limits>

using namespace std;

using ll = long long;
using pll = pair<ll, ll>;
using vl = vector<ll>;
using vll = vector<vl>;
using vpll = vector<pll>;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rrep(i, n) for (int i = ((int)(n)-1); i >= 0; i--)
#define Rep(i, n) for (long long i = 0; i< (long long)(n); i++)
#define RRep(i,n) for (long long i = ((long long)(n)-1); i>=0; i--)
#define all(v) (v).begin(),(v).end()

struct Edge{
    int to;
    int weight;
    Edge(int t, int w) : to(t), weight(w){ }
};

struct edge{
    int from;
    int to;
    int weight;
    edge(int f,int t,int w) : from(f), to(t), weight(w){ }
};

using Graph = vector<vector<Edge>>;
using graph = vector<vector<int>>;
using edges = vector<edge>;

template <class T>
inline bool chmin(T &a, T b){
    if (a > b){
        a = b;
        return true;
    }
    return false;
}

template <class T>
inline bool chmax(T &a, T b){
    if (a < b){
        a = b;
        return true;
    }
    return false;
}

template <class T>
T gcd(T a, T b){
    if( b==0 ) return a;
    return gcd(b, a%b);
}

template <class T>
T lcm(T a, T b){
    return (a*b)/gcd(a,b);
}

constexpr ll INF = 1e9+7;

vector<int> dx = {-1,0,1,0};
vector<int> dy = {0,-1,0,1};

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int 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;
      string s = "";
      cout << 11 << endl;
      rep(i,n-2){
        s += "0";
        cout << "1" << s << "1" << endl; 
      }
    }
}
