#include <bits/stdc++.h>
const int N = 101;
int n, p, i, j;
bool f[N][N];
int main() {
  scanf("%d", &n);
  for (p = 3; p * (p - 1) * (p - 2) / 6 <= n; p++)
    ;
  p--;
  n -= p * (p - 1) * (p - 2) / 6;
  for (i = 1; i <= p; i++)
    for (j = 1; j <= p; j++)
      if (i != j) f[i][j] = 1;
  for (i = p; i > 1; i--)
    if (i * (i - 1) / 2 <= n) {
      n -= i * (i - 1) / 2;
      p++;
      for (j = 1; j <= i; j++) f[p][j] = f[j][p] = 1;
      i++;
    }
  printf("%d\n", p);
  for (i = 1; i <= p; i++, puts(""))
    for (j = 1; j <= p; j++) printf("%d", f[i][j]);
}

### Prompt

Develop a solution in Cpp 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>
const int N = 101;
int n, p, i, j;
bool f[N][N];
int main() {
  scanf("%d", &n);
  for (p = 3; p * (p - 1) * (p - 2) / 6 <= n; p++)
    ;
  p--;
  n -= p * (p - 1) * (p - 2) / 6;
  for (i = 1; i <= p; i++)
    for (j = 1; j <= p; j++)
      if (i != j) f[i][j] = 1;
  for (i = p; i > 1; i--)
    if (i * (i - 1) / 2 <= n) {
      n -= i * (i - 1) / 2;
      p++;
      for (j = 1; j <= i; j++) f[p][j] = f[j][p] = 1;
      i++;
    }
  printf("%d\n", p);
  for (i = 1; i <= p; i++, puts(""))
    for (j = 1; j <= p; j++) printf("%d", f[i][j]);
}

```