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
//https://onlinejudge.u-aizu.ac.jp/beta/room.html#KUPC2020Spring/problems/M
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#define all(x) (x).begin(),(x).end()
//#pragma GCC optimize ("-O3")
using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
typedef long long ll; const int inf = INT_MAX / 2; const ll infl = 1LL << 60;
template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
//---------------------------------------------------------------------------------------------------
/*---------------------------------------------------------------------------------------------------
　　　　　　　　　　　 ∧＿∧
　　　　　 ∧＿∧ 　（´<_｀ ）　 Welcome to My Coding Space!
　　　　 （ ´_ゝ`）　/　 ⌒i     @hamayanhamayan0
　　　　／　　　＼　 　  |　|
　　　 /　　 /￣￣￣￣/　　|
　 ＿_(__ﾆつ/　    ＿/ .| .|＿＿＿＿
　 　　　＼/＿＿＿＿/　（u　⊃
---------------------------------------------------------------------------------------------------*/














int N;
//---------------------------------------------------------------------------------------------------
void _main() {
	cin >> N;

	vector<string> ans;
	if (N == 1) ans.push_back("0");
	else if (N == 2) ans.push_back("0"), ans.push_back("11");
	else {
		rep(i, 0, N - 1) {
			string ones = "";
			rep(j, 0, i) ones += "1";
			ans.push_back("0" + ones + "0");
		}
	}

	int n = ans.size();
	printf("%d\n", n);
	fore(s, ans) printf("%s\n", s.c_str());
}





/* ///////////////////////// writeup1 start
///////////////////////// writeup2 start
N=1 {0}
N=2 {0, 11}
N=3 {00, 111}
N=4 {00, 111, 1010}
N=5 {00, 010, 0110, 01110}
N=6 {101010, 10100, 1100}
///////////////////////// writeup2 end */
//　
