#include <bits/stdc++.h>
using namespace std;
int k, a[110][110], n = 0, d[110], j, l, i;
int main() {
  scanf("%d", &k);
  a[1][2] = 1;
  a[2][1] = 1;
  d[3] = 1;
  for (i = 4; i <= 100; i++) d[i] = d[i - 1] + (i - 2);
  i = 2;
  while (k > 0) {
    i++;
    if (d[i] <= k) {
      k -= d[i];
      for (j = 1; j <= i - 1; j++) a[i][j] = 1, a[j][i] = 1;
    } else {
      j = i;
      while (d[j] > k) j--;
      k -= d[j];
      for (j = j - 1; j >= 1; j--) a[i][j] = 1, a[j][i] = 1;
    }
  }
  printf("%d\n", i);
  for (j = 1; j <= i; j++) {
    for (l = 1; l <= i; l++) printf("%d", a[j][l]);
    printf("\n");
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 k, a[110][110], n = 0, d[110], j, l, i;
int main() {
  scanf("%d", &k);
  a[1][2] = 1;
  a[2][1] = 1;
  d[3] = 1;
  for (i = 4; i <= 100; i++) d[i] = d[i - 1] + (i - 2);
  i = 2;
  while (k > 0) {
    i++;
    if (d[i] <= k) {
      k -= d[i];
      for (j = 1; j <= i - 1; j++) a[i][j] = 1, a[j][i] = 1;
    } else {
      j = i;
      while (d[j] > k) j--;
      k -= d[j];
      for (j = j - 1; j >= 1; j--) a[i][j] = 1, a[j][i] = 1;
    }
  }
  printf("%d\n", i);
  for (j = 1; j <= i; j++) {
    for (l = 1; l <= i; l++) printf("%d", a[j][l]);
    printf("\n");
  }
  return 0;
}

```