#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 1e2 + 7;
int G[MAX_N][MAX_N];
void link(int a, int b) { G[a][b] = G[b][a] = true; }
void check(int des, int n) {
  int ret = 0;
  for (int i = (1); i < (n + 1); i++)
    for (int j = (1); j < (n + 1); j++)
      for (int k = (1); k < (n + 1); k++)
        if (G[i][j] && G[i][k] && G[j][k]) ret++;
  ret /= 6;
  cout << "ret"
       << " => " << ret << endl;
  cout << "des"
       << " => " << des << endl;
  puts(ret == des ? "AC" : "WA");
}
int C2[MAX_N];
int main() {
  for (int i = (2); i < (MAX_N); i++) C2[i] = i * (i - 1) / 2;
  link(1, 2);
  int n;
  cin >> n;
  int _n = n;
  int base = 2;
  for (; n >= C2[base]; base++) {
    n -= C2[base];
    for (int i = (1); i < (base + 1); i++) link(i, base + 1);
  }
  for (; n;) {
    int pos = lower_bound(C2, C2 + MAX_N, n + 1) - C2;
    pos--;
    n -= C2[pos];
    ++base;
    for (int i = (1); i < (pos + 1); i++) link(base, i);
  }
  printf("%d\n", base);
  for (int i = (1); i < (base + 1); i++) {
    for (int j = (1); j < (base + 1); j++) printf("%d", G[i][j]);
    printf("\n");
  }
}

### 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>
using namespace std;
const int MAX_N = 1e2 + 7;
int G[MAX_N][MAX_N];
void link(int a, int b) { G[a][b] = G[b][a] = true; }
void check(int des, int n) {
  int ret = 0;
  for (int i = (1); i < (n + 1); i++)
    for (int j = (1); j < (n + 1); j++)
      for (int k = (1); k < (n + 1); k++)
        if (G[i][j] && G[i][k] && G[j][k]) ret++;
  ret /= 6;
  cout << "ret"
       << " => " << ret << endl;
  cout << "des"
       << " => " << des << endl;
  puts(ret == des ? "AC" : "WA");
}
int C2[MAX_N];
int main() {
  for (int i = (2); i < (MAX_N); i++) C2[i] = i * (i - 1) / 2;
  link(1, 2);
  int n;
  cin >> n;
  int _n = n;
  int base = 2;
  for (; n >= C2[base]; base++) {
    n -= C2[base];
    for (int i = (1); i < (base + 1); i++) link(i, base + 1);
  }
  for (; n;) {
    int pos = lower_bound(C2, C2 + MAX_N, n + 1) - C2;
    pos--;
    n -= C2[pos];
    ++base;
    for (int i = (1); i < (pos + 1); i++) link(base, i);
  }
  printf("%d\n", base);
  for (int i = (1); i < (base + 1); i++) {
    for (int j = (1); j < (base + 1); j++) printf("%d", G[i][j]);
    printf("\n");
  }
}

```