#include <bits/stdc++.h>
using namespace std;
int mp[111][111];
int main() {
  int k;
  memset(mp, 0, sizeof(mp));
  scanf("%d", &k);
  int n = 3;
  while (n * (n - 1) * (n - 2) / 6 < k) {
    n++;
  }
  if (n * (n - 1) * (n - 2) / 6 == k) {
    cout << n << endl;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        if (i == j)
          printf("0");
        else
          printf("1");
      }
      puts("");
    }
    return 0;
  }
  n--;
  k -= n * (n - 1) * (n - 2) / 6;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i != j) mp[i][j] = 1;
    }
  }
  while (k) {
    n++;
    int tp = 2;
    while (tp * (tp - 1) / 2 <= k) {
      tp++;
    }
    tp--;
    k -= tp * (tp - 1) / 2;
    for (int i = 1; i <= tp; i++) {
      mp[n][i] = 1;
      mp[i][n] = 1;
    }
  }
  cout << n << endl;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      printf("%d", mp[i][j]);
    }
    puts("");
  }
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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>
using namespace std;
int mp[111][111];
int main() {
  int k;
  memset(mp, 0, sizeof(mp));
  scanf("%d", &k);
  int n = 3;
  while (n * (n - 1) * (n - 2) / 6 < k) {
    n++;
  }
  if (n * (n - 1) * (n - 2) / 6 == k) {
    cout << n << endl;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        if (i == j)
          printf("0");
        else
          printf("1");
      }
      puts("");
    }
    return 0;
  }
  n--;
  k -= n * (n - 1) * (n - 2) / 6;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i != j) mp[i][j] = 1;
    }
  }
  while (k) {
    n++;
    int tp = 2;
    while (tp * (tp - 1) / 2 <= k) {
      tp++;
    }
    tp--;
    k -= tp * (tp - 1) / 2;
    for (int i = 1; i <= tp; i++) {
      mp[n][i] = 1;
      mp[i][n] = 1;
    }
  }
  cout << n << endl;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      printf("%d", mp[i][j]);
    }
    puts("");
  }
  return 0;
}

```