#include <bits/stdc++.h>
using namespace std;
char G[100][100];
int main() {
  ios_base::sync_with_stdio(false);
  int k;
  scanf("%d", &k);
  int n = 3;
  while (n * (n - 1) * (n - 2) / 6 <= k) {
    n++;
  }
  for (int i = 0; i < 100; ++i)
    for (int j = 0; j < 100; ++j) G[i][j] = '0';
  for (int i = 0; i < n - 1; i++) {
    for (int j = i + 1; j < n - 1; j++) {
      G[i][j] = '1';
      G[j][i] = '1';
    }
  }
  k -= ((n - 1) * (n - 2) * (n - 3) / 6);
  int u = n - 1;
  while (k >= 1) {
    int v = 3;
    while (v * (v - 1) / 2 <= k) v++;
    for (int w = 0; w < v - 1; ++w) {
      G[u][w] = '1';
      G[w][u] = '1';
    }
    k -= (v - 1) * (v - 2) / 2;
    u++;
  }
  printf("100\n");
  for (int i = 0; i < 100; ++i) {
    for (int j = 0; j < 100; ++j) printf("%c", G[i][j]);
    printf("\n");
  }
  return 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>
using namespace std;
char G[100][100];
int main() {
  ios_base::sync_with_stdio(false);
  int k;
  scanf("%d", &k);
  int n = 3;
  while (n * (n - 1) * (n - 2) / 6 <= k) {
    n++;
  }
  for (int i = 0; i < 100; ++i)
    for (int j = 0; j < 100; ++j) G[i][j] = '0';
  for (int i = 0; i < n - 1; i++) {
    for (int j = i + 1; j < n - 1; j++) {
      G[i][j] = '1';
      G[j][i] = '1';
    }
  }
  k -= ((n - 1) * (n - 2) * (n - 3) / 6);
  int u = n - 1;
  while (k >= 1) {
    int v = 3;
    while (v * (v - 1) / 2 <= k) v++;
    for (int w = 0; w < v - 1; ++w) {
      G[u][w] = '1';
      G[w][u] = '1';
    }
    k -= (v - 1) * (v - 2) / 2;
    u++;
  }
  printf("100\n");
  for (int i = 0; i < 100; ++i) {
    for (int j = 0; j < 100; ++j) printf("%c", G[i][j]);
    printf("\n");
  }
  return 0;
}

```