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;

#define ll long long

/*
5
000000
00001
1000
011
11
*/

/*
7
0000000000
000000001
10000000
0000011
110000
00111
1110
101
*/



signed main(){
   int n;
   vector<string > ans;
   cin >> n;
   if (n == 1){
      cout << 1 << endl;
      cout << "0" << endl;
      return 0;
   }
   if (n == 2) { 
      cout << 2 << endl;
      cout << "1" << endl;
      cout << "00" << endl;
      return 0;
   }
   int num0 = n - 2;
   while(num0 >= 0){
      ans.push_back("1" + string(num0, '0') + "1");
      num0 -= 1;
   }

   /*if (n % 3 == 0){
      cout << ans.size() - 1 << endl;
      for(int i = 0; i < ans.size() - 1; i += 1){
         cout << ans[i] << endl;
      }
   }*/
   //else {
   cout << ans.size() << endl;
   for(int i = 0; i < ans.size(); i += 1){
      cout << ans[i] << endl;
   }
   //}
}
