#include <bits/stdc++.h>

using namespace std;

int n;
string a[1000];

int depth[1000];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] = "#" + a[i];
  }

  int mx = -1;
  for (int start = 1; start <= n; start++) {
    queue<int> q;
    for (int i = 1; i <= n; ++i) {
      depth[i] = 0;
    }
    q.push(start);
    depth[start] = 1;
    bool ok = true;
    while (!q.empty()) {
      int v = q.front();
      q.pop();
      for (int j = 1; j <= n; j++) {
        if (a[v][j] == '1') {
          if (depth[j] == 0) {
            q.push(j);
            depth[j] = depth[v] + 1;
          } else if (depth[j] != depth[v] + 1 && depth[j] != depth[v] - 1) {
            ok = false;
          }
        }
      }
    }
    if (ok) {
      for (int j = 1; j<= n; j++) {
        mx = max(mx, depth[j]);
      }
    }
  }
  cout << mx << "\n";
  return 0;
}
### Prompt

In Cpp, your task is to solve 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 n;
string a[1000];

int depth[1000];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] = "#" + a[i];
  }

  int mx = -1;
  for (int start = 1; start <= n; start++) {
    queue<int> q;
    for (int i = 1; i <= n; ++i) {
      depth[i] = 0;
    }
    q.push(start);
    depth[start] = 1;
    bool ok = true;
    while (!q.empty()) {
      int v = q.front();
      q.pop();
      for (int j = 1; j <= n; j++) {
        if (a[v][j] == '1') {
          if (depth[j] == 0) {
            q.push(j);
            depth[j] = depth[v] + 1;
          } else if (depth[j] != depth[v] + 1 && depth[j] != depth[v] - 1) {
            ok = false;
          }
        }
      }
    }
    if (ok) {
      for (int j = 1; j<= n; j++) {
        mx = max(mx, depth[j]);
      }
    }
  }
  cout << mx << "\n";
  return 0;
}
```