#include <bits/stdc++.h>
#define DEBUG if(0)
#define lli long long int
#define ldouble long double
using namespace std;

const int maxN = 200; int n;
int dist[maxN];
vector<int> graph[maxN];

bool bfs(int origin)
{
  memset(dist, -1, sizeof(dist));
  queue<int> q; q.push(origin), dist[origin] = 0;
  while (!q.empty())
  {
    int u = q.front(); q.pop();
    for (int v: graph[u])
      if (dist[v] == -1)
        q.push(v), dist[v] = dist[u] + 1;
      else if (dist[u] == dist[v])
        return false;
  }
  return true;
}

int main()
{
  scanf("%d", &n);
  for (int i = 0; i < n; i ++)
    for (int j = 0; j < n; j ++)
    {
      int has; scanf("%1d", &has);
      if (has) graph[i].push_back(j);
    }

  bool can = true;
  int ans = 0;
  for (int i = 0; i < n && can; i ++)
  {
    can = bfs(i);
    ans = max(ans, *max_element(dist, dist+n) + 1);
  }
  if (can) bfs(max_element(dist, dist+n) - dist);
  printf("%d\n", !can ? -1 : ans);

  return 0;
}
### Prompt

Your task is to create a cpp solution to the following problem:

Given is a connected undirected graph with N vertices and M edges. The vertices are numbered 1 to N, and the edges are described by a grid of characters S. If S_{i,j} is `1`, there is an edge connecting Vertex i and j; otherwise, there is no such edge.

Determine whether it is possible to divide the vertices into non-empty sets V_1, \dots, V_k such that the following condition is satisfied. If the answer is yes, find the maximum possible number of sets, k, in such a division.

* Every edge connects two vertices belonging to two "adjacent" sets. More formally, for every edge (i,j), there exists 1\leq t\leq k-1 such that i\in V_t,j\in V_{t+1} or i\in V_{t+1},j\in V_t holds.

Constraints

* 2 \leq N \leq 200
* S_{i,j} is `0` or `1`.
* S_{i,i} is `0`.
* S_{i,j}=S_{j,i}
* The given graph is connected.
* N is an integer.

Input

Input is given from Standard Input in the following format:


N
S_{1,1}...S_{1,N}
:
S_{N,1}...S_{N,N}


Output

If it is impossible to divide the vertices into sets so that the condition is satisfied, print -1. Otherwise, print the maximum possible number of sets, k, in a division that satisfies the condition.

Examples

Input

2
01
10


Output

2


Input

3
011
101
110


Output

-1


Input

6
010110
101001
010100
101000
100000
010000


Output

4

### Response

```cpp
#include <bits/stdc++.h>
#define DEBUG if(0)
#define lli long long int
#define ldouble long double
using namespace std;

const int maxN = 200; int n;
int dist[maxN];
vector<int> graph[maxN];

bool bfs(int origin)
{
  memset(dist, -1, sizeof(dist));
  queue<int> q; q.push(origin), dist[origin] = 0;
  while (!q.empty())
  {
    int u = q.front(); q.pop();
    for (int v: graph[u])
      if (dist[v] == -1)
        q.push(v), dist[v] = dist[u] + 1;
      else if (dist[u] == dist[v])
        return false;
  }
  return true;
}

int main()
{
  scanf("%d", &n);
  for (int i = 0; i < n; i ++)
    for (int j = 0; j < n; j ++)
    {
      int has; scanf("%1d", &has);
      if (has) graph[i].push_back(j);
    }

  bool can = true;
  int ans = 0;
  for (int i = 0; i < n && can; i ++)
  {
    can = bfs(i);
    ans = max(ans, *max_element(dist, dist+n) + 1);
  }
  if (can) bfs(max_element(dist, dist+n) - dist);
  printf("%d\n", !can ? -1 : ans);

  return 0;
}
```