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

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

  int r = 0;
  for (int start = 0; start < n; start++) {
    bool b = true;
    vector<int> d(n, -1);
    queue<int> q;
    q.push(start);
    d[start] = 0;
    while (b && !q.empty()) {
      int i = q.front();
      q.pop();
      for (auto j : v[i]) {
        if (d[j] == -1) {
          d[j] = d[i] + 1;
          q.push(j);
        } else if (d[j] != d[i] - 1 && d[j] != d[i] + 1) {
          b = false;
          break;
        }
      }
    }
    if (b) r = max(r, *max_element(d.begin(), d.end()) + 1);
  }
  cout << (r ? r : -1) << endl;
}

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

  int r = 0;
  for (int start = 0; start < n; start++) {
    bool b = true;
    vector<int> d(n, -1);
    queue<int> q;
    q.push(start);
    d[start] = 0;
    while (b && !q.empty()) {
      int i = q.front();
      q.pop();
      for (auto j : v[i]) {
        if (d[j] == -1) {
          d[j] = d[i] + 1;
          q.push(j);
        } else if (d[j] != d[i] - 1 && d[j] != d[i] + 1) {
          b = false;
          break;
        }
      }
    }
    if (b) r = max(r, *max_element(d.begin(), d.end()) + 1);
  }
  cout << (r ? r : -1) << endl;
}

```