#include <bits/stdc++.h>

using namespace std;

int n, lvl[201]; char c;
vector<int> adj[201];

int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int k = 0; k < n; k++) {
      cin >> c;
      if (c == '1') {
        adj[i].push_back(k);
      }
    }
  }
  int res = 0;
  for (int i = 0; i < n; i++) {
    memset(lvl, -1, sizeof lvl);
    queue<int> nds;
    lvl[i] = 0;
    nds.push(i);
    bool ispos = true;
    int maxlvl = 0;
    while (!nds.empty()) {
      int i = nds.front(); nds.pop();
      for (int j : adj[i]) {
        if (lvl[j] == -1) {
          lvl[j] = lvl[i] + 1;
          maxlvl = max(maxlvl, lvl[j]);
          nds.push(j);
        } else if (lvl[j] != lvl[i] + 1 && lvl[j] != lvl[i] - 1) {
          ispos = false;
          break;
        }
      }
    }
    if (!ispos) {
      cout << -1 << endl;
      return 0;
    }
    res = max(res, maxlvl + 1);
  }
  cout << res << endl;
  return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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, lvl[201]; char c;
vector<int> adj[201];

int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int k = 0; k < n; k++) {
      cin >> c;
      if (c == '1') {
        adj[i].push_back(k);
      }
    }
  }
  int res = 0;
  for (int i = 0; i < n; i++) {
    memset(lvl, -1, sizeof lvl);
    queue<int> nds;
    lvl[i] = 0;
    nds.push(i);
    bool ispos = true;
    int maxlvl = 0;
    while (!nds.empty()) {
      int i = nds.front(); nds.pop();
      for (int j : adj[i]) {
        if (lvl[j] == -1) {
          lvl[j] = lvl[i] + 1;
          maxlvl = max(maxlvl, lvl[j]);
          nds.push(j);
        } else if (lvl[j] != lvl[i] + 1 && lvl[j] != lvl[i] - 1) {
          ispos = false;
          break;
        }
      }
    }
    if (!ispos) {
      cout << -1 << endl;
      return 0;
    }
    res = max(res, maxlvl + 1);
  }
  cout << res << endl;
  return 0;
}
```