#include <bits/stdc++.h>
using namespace std;

int main() {
  const int inf = 100000000;
  int n;
  cin >> n;
  vector<string> s(n);
  for (int i = 0; i < n; i++) cin >> s.at(i);
  vector<vector<int>> cost(n, vector<int>(n, inf));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) {
        cost.at(i).at(j) = 0;
      } else if (s.at(i).at(j) == '1') {
        cost.at(i).at(j) = 1;
      }
    }
  }
  for (int k = 0; k < n; k++) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        int d0 = cost.at(i).at(j), d1 = cost.at(i).at(k) + cost.at(k).at(j);
        if (d0 < inf && d1 < inf && abs(d1 - d0) % 2 == 1) {
          cout << -1 << endl;
          return 0;
        }
        cost.at(i).at(j) = min(d0, d1);
      }
    }
  }
  int dmax = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      dmax = max(dmax, cost.at(i).at(j));
    }
  }
  cout << dmax + 1 << endl;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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() {
  const int inf = 100000000;
  int n;
  cin >> n;
  vector<string> s(n);
  for (int i = 0; i < n; i++) cin >> s.at(i);
  vector<vector<int>> cost(n, vector<int>(n, inf));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) {
        cost.at(i).at(j) = 0;
      } else if (s.at(i).at(j) == '1') {
        cost.at(i).at(j) = 1;
      }
    }
  }
  for (int k = 0; k < n; k++) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        int d0 = cost.at(i).at(j), d1 = cost.at(i).at(k) + cost.at(k).at(j);
        if (d0 < inf && d1 < inf && abs(d1 - d0) % 2 == 1) {
          cout << -1 << endl;
          return 0;
        }
        cost.at(i).at(j) = min(d0, d1);
      }
    }
  }
  int dmax = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      dmax = max(dmax, cost.at(i).at(j));
    }
  }
  cout << dmax + 1 << endl;
}
```