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

const int maxN = 200, inf = 1e6; int n;
int edge[maxN][maxN], dist[maxN][maxN];

int floydWarshall()
{
  for (int k = 0; k < n; k ++)
    for (int u = 0; u < n; u ++)
      for (int v = 0; v < n; v ++)
      {
        if (edge[u][v] != 0 && dist[k][u] + dist[k][v] < inf && dist[k][u] == dist[k][v])
          return -1;
        dist[u][v] = min(dist[u][v], dist[u][k] + dist[k][v]);
      }
  int ans = 0;
  for (int i = 0; i < n; i ++) ans = max(ans, *max_element(dist[i], dist[i]+n) + 1);
  return ans;
}

int main()
{
  scanf("%d", &n);
  for (int i = 0; i < n; i ++)
    for (int j = 0; j < n; j ++)
    {
      scanf("%1d", &edge[i][j]);
      dist[i][j] = edge[i][j] ? 1 : (i != j) *inf;
    }

  int ans = floydWarshall();
  printf("%d\n", ans);

  return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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, inf = 1e6; int n;
int edge[maxN][maxN], dist[maxN][maxN];

int floydWarshall()
{
  for (int k = 0; k < n; k ++)
    for (int u = 0; u < n; u ++)
      for (int v = 0; v < n; v ++)
      {
        if (edge[u][v] != 0 && dist[k][u] + dist[k][v] < inf && dist[k][u] == dist[k][v])
          return -1;
        dist[u][v] = min(dist[u][v], dist[u][k] + dist[k][v]);
      }
  int ans = 0;
  for (int i = 0; i < n; i ++) ans = max(ans, *max_element(dist[i], dist[i]+n) + 1);
  return ans;
}

int main()
{
  scanf("%d", &n);
  for (int i = 0; i < n; i ++)
    for (int j = 0; j < n; j ++)
    {
      scanf("%1d", &edge[i][j]);
      dist[i][j] = edge[i][j] ? 1 : (i != j) *inf;
    }

  int ans = floydWarshall();
  printf("%d\n", ans);

  return 0;
}
```