#include <bits/stdc++.h>
#pragma comment(linker, "/stack:225450978")
#pragma GCC optimize("Ofast")
using namespace std;
const long long Mod = 1000000007LL, INF = 1e9, LINF = 1e18;
const long double Pi = 3.141592653589793116, EPS = 1e-9,
                  Gold = ((1 + sqrt(5)) / 2);
long long keymod[] = {1000000007LL, 1000000009LL, 1000000021LL, 1000000033LL};
long long keyCount = sizeof(keymod) / sizeof(long long);
mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng64(chrono::steady_clock::now().time_since_epoch().count());
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcountll(s);
}
auto TimeStart = chrono::steady_clock::now();
auto TimeEnd = chrono::steady_clock::now();
void ControlIO(int argc, char* argv[]);
void TimerStart();
void TimerStop();
void Exit();
string cppstr_infile = "FILE.IN";
string cppstr_outfile = "FILE.OUT";
long long k;
vector<vector<char> > Mat(100, vector<char>(100, '0'));
long long TripletFull_max(long long r) {
  long long top = 0, bot = 100, ans = -1;
  while (top <= bot) {
    long long mid = (top + bot) / 2;
    if (mid * (mid - 1) * (mid - 2) / 6 <= r) {
      ans = mid;
      top = mid + 1;
    } else
      bot = mid - 1;
  }
  return ans;
}
void pairwiseFill(long long L, long long R) {
  for (long long i = L; i <= R; i++) {
    for (long long j = L; j <= R; j++) {
      if (i != j) Mat[i][j] = '1';
    }
  }
}
long long DoubletFull_max(long long r) {
  long long top = 0, bot = 100, ans = -1;
  while (top <= bot) {
    long long mid = (top + bot) / 2;
    if (mid * (mid - 1) / 2 <= r) {
      ans = mid;
      top = mid + 1;
    } else
      bot = mid - 1;
  }
  return ans;
}
void bridgeFill(long long src, long long L, long long R) {
  for (long long i = L; i <= R; i++) {
    Mat[src][i] = '1';
    Mat[i][src] = '1';
  }
}
void Input() { cin >> k; }
void Solve() {
  long long n = TripletFull_max(k);
  pairwiseFill(0, n - 1);
  k -= n * (n - 1) * (n - 2) / 6;
  while (k > 0) {
    long long z = DoubletFull_max(k);
    bridgeFill(n, 0, z - 1);
    n++;
    k -= z * (z - 1) / 2;
  }
  cout << n << '\n';
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      cout << Mat[i][j];
    }
    cout << '\n';
  }
}
int main(int argc, char* argv[]) {
  ControlIO(argc, argv);
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  Input();
  TimerStart();
  Solve();
  TimerStop();
  return 0;
}
void ControlIO(int argc, char* argv[]) {
  char* infile = new char[cppstr_infile.size() + 1];
  char* outfile = new char[cppstr_outfile.size() + 1];
  strcpy(infile, cppstr_infile.c_str());
  strcpy(outfile, cppstr_outfile.c_str());
}
void TimerStart() {}
void TimerStop() {}
void Exit() {
  TimerStop();
  exit(0);
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

John Doe started thinking about graphs. After some thought he decided that he wants to paint an undirected graph, containing exactly k cycles of length 3. 

A cycle of length 3 is an unordered group of three distinct graph vertices a, b and c, such that each pair of them is connected by a graph edge. 

John has been painting for long, but he has not been a success. Help him find such graph. Note that the number of vertices there shouldn't exceed 100, or else John will have problems painting it.

Input

A single line contains an integer k (1 ≤ k ≤ 105) — the number of cycles of length 3 in the required graph.

Output

In the first line print integer n (3 ≤ n ≤ 100) — the number of vertices in the found graph. In each of next n lines print n characters "0" and "1": the i-th character of the j-th line should equal "0", if vertices i and j do not have an edge between them, otherwise it should equal "1". Note that as the required graph is undirected, the i-th character of the j-th line must equal the j-th character of the i-th line. The graph shouldn't contain self-loops, so the i-th character of the i-th line must equal "0" for all i.

Examples

Input

1


Output

3
011
101
110


Input

10


Output

5
01111
10111
11011
11101
11110

### Response

```cpp
#include <bits/stdc++.h>
#pragma comment(linker, "/stack:225450978")
#pragma GCC optimize("Ofast")
using namespace std;
const long long Mod = 1000000007LL, INF = 1e9, LINF = 1e18;
const long double Pi = 3.141592653589793116, EPS = 1e-9,
                  Gold = ((1 + sqrt(5)) / 2);
long long keymod[] = {1000000007LL, 1000000009LL, 1000000021LL, 1000000033LL};
long long keyCount = sizeof(keymod) / sizeof(long long);
mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng64(chrono::steady_clock::now().time_since_epoch().count());
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcountll(s);
}
auto TimeStart = chrono::steady_clock::now();
auto TimeEnd = chrono::steady_clock::now();
void ControlIO(int argc, char* argv[]);
void TimerStart();
void TimerStop();
void Exit();
string cppstr_infile = "FILE.IN";
string cppstr_outfile = "FILE.OUT";
long long k;
vector<vector<char> > Mat(100, vector<char>(100, '0'));
long long TripletFull_max(long long r) {
  long long top = 0, bot = 100, ans = -1;
  while (top <= bot) {
    long long mid = (top + bot) / 2;
    if (mid * (mid - 1) * (mid - 2) / 6 <= r) {
      ans = mid;
      top = mid + 1;
    } else
      bot = mid - 1;
  }
  return ans;
}
void pairwiseFill(long long L, long long R) {
  for (long long i = L; i <= R; i++) {
    for (long long j = L; j <= R; j++) {
      if (i != j) Mat[i][j] = '1';
    }
  }
}
long long DoubletFull_max(long long r) {
  long long top = 0, bot = 100, ans = -1;
  while (top <= bot) {
    long long mid = (top + bot) / 2;
    if (mid * (mid - 1) / 2 <= r) {
      ans = mid;
      top = mid + 1;
    } else
      bot = mid - 1;
  }
  return ans;
}
void bridgeFill(long long src, long long L, long long R) {
  for (long long i = L; i <= R; i++) {
    Mat[src][i] = '1';
    Mat[i][src] = '1';
  }
}
void Input() { cin >> k; }
void Solve() {
  long long n = TripletFull_max(k);
  pairwiseFill(0, n - 1);
  k -= n * (n - 1) * (n - 2) / 6;
  while (k > 0) {
    long long z = DoubletFull_max(k);
    bridgeFill(n, 0, z - 1);
    n++;
    k -= z * (z - 1) / 2;
  }
  cout << n << '\n';
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      cout << Mat[i][j];
    }
    cout << '\n';
  }
}
int main(int argc, char* argv[]) {
  ControlIO(argc, argv);
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  Input();
  TimerStart();
  Solve();
  TimerStop();
  return 0;
}
void ControlIO(int argc, char* argv[]) {
  char* infile = new char[cppstr_infile.size() + 1];
  char* outfile = new char[cppstr_outfile.size() + 1];
  strcpy(infile, cppstr_infile.c_str());
  strcpy(outfile, cppstr_outfile.c_str());
}
void TimerStart() {}
void TimerStop() {}
void Exit() {
  TimerStop();
  exit(0);
}

```