#include <bits/stdc++.h>
 
using namespace std;
 
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<string> g(n);
  for (int i = 0; i < n; i++) {
    cin >> g[i];
  }
  int ans = 0;
  for (int st = 0; st < n; st++) {
    vector<int> dist(n, -1);
    vector<int> que(1, st);
    dist[st] = 0;
    for (int b = 0; b < (int) que.size(); b++) {
      for (int i = 0; i < n; i++) {
        if (g[que[b]][i] == '1' && dist[i] == -1) {
          que.push_back(i);
          dist[i] = dist[que[b]] + 1;
        }
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if (g[i][j] == '1' && dist[i] == dist[j]) {
          cout << -1 << '\n';
          return 0;
        }
      }
    }
    int mx = *max_element(dist.begin(), dist.end());
    ans = max(ans, mx);
  }
  cout << ans + 1 << endl;
  return 0;
}
### Prompt

Create a solution in CPP for 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>
 
using namespace std;
 
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<string> g(n);
  for (int i = 0; i < n; i++) {
    cin >> g[i];
  }
  int ans = 0;
  for (int st = 0; st < n; st++) {
    vector<int> dist(n, -1);
    vector<int> que(1, st);
    dist[st] = 0;
    for (int b = 0; b < (int) que.size(); b++) {
      for (int i = 0; i < n; i++) {
        if (g[que[b]][i] == '1' && dist[i] == -1) {
          que.push_back(i);
          dist[i] = dist[que[b]] + 1;
        }
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if (g[i][j] == '1' && dist[i] == dist[j]) {
          cout << -1 << '\n';
          return 0;
        }
      }
    }
    int mx = *max_element(dist.begin(), dist.end());
    ans = max(ans, mx);
  }
  cout << ans + 1 << endl;
  return 0;
}
```