#include <bits/stdc++.h>

using namespace std;

int main() {
  int n;
  cin >> n;
  vector< vector<int> > g(n);
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    for (int j = 0; j < n; j++) {
      if (s[j] == '1') {
        g[i].push_back(j);
      }
    }
  }

  int ans = -1;
  for (int i = 0; i < n; i++) {
    vector<int> v(n, -1);
    v[i] = 0;
    bool ok = true;
    int num = 0;
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        if (v[k] == j) {
          for (auto u : g[k]) {
            if (v[u] == -1) {
              v[u] = num = j + 1;
            } else if (abs(v[u] - j) != 1) {
              ok = false;
            }
          }
        }
      }
    }
    // for (auto x : v) {
    //   cout << x << ' ';
    // }
    // cout << '\n';
    if (ok) ans = max(ans, num + 1);
  }
  cout << ans << '\n';

  return 0;
}
### Prompt

Your challenge is to write 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>

using namespace std;

int main() {
  int n;
  cin >> n;
  vector< vector<int> > g(n);
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    for (int j = 0; j < n; j++) {
      if (s[j] == '1') {
        g[i].push_back(j);
      }
    }
  }

  int ans = -1;
  for (int i = 0; i < n; i++) {
    vector<int> v(n, -1);
    v[i] = 0;
    bool ok = true;
    int num = 0;
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        if (v[k] == j) {
          for (auto u : g[k]) {
            if (v[u] == -1) {
              v[u] = num = j + 1;
            } else if (abs(v[u] - j) != 1) {
              ok = false;
            }
          }
        }
      }
    }
    // for (auto x : v) {
    //   cout << x << ' ';
    // }
    // cout << '\n';
    if (ok) ans = max(ans, num + 1);
  }
  cout << ans << '\n';

  return 0;
}
```