#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;

signed main() {
#ifdef LOCAL
  assert(freopen("b.in", "r", stdin));
#endif
  int n;
  cin >> n;
  vector<string> g(n);
  for (int i = 0; i < n; ++i) {
    cin >> g[i];
  }
  int res = -1;
  for (int s = 0; s < n; ++s) {
    vector<int> q{s};
    vector<int> dist(n, -1);
    dist[s] = 0;
    bool fail = false;
    for (int ii = 0; ii < (int) q.size(); ++ii) {
      int u = q[ii];
      for (int v = 0; v < n; ++v) {
        if (g[u][v] == '0') {
          continue;
        }
        if (dist[v] != -1) {
          if (abs(dist[v] - dist[u]) != 1) {
            fail = true;
          }
          continue;
        }
        dist[v] = dist[u] + 1;
        q.push_back(v);
      }
    }
    if (!fail) {
      res = max(res, *max_element(dist.begin(), dist.end()) + 1);
    }
  }
  cout << res << '\n';
}

### 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;
using ll = long long;
using ld = long double;

signed main() {
#ifdef LOCAL
  assert(freopen("b.in", "r", stdin));
#endif
  int n;
  cin >> n;
  vector<string> g(n);
  for (int i = 0; i < n; ++i) {
    cin >> g[i];
  }
  int res = -1;
  for (int s = 0; s < n; ++s) {
    vector<int> q{s};
    vector<int> dist(n, -1);
    dist[s] = 0;
    bool fail = false;
    for (int ii = 0; ii < (int) q.size(); ++ii) {
      int u = q[ii];
      for (int v = 0; v < n; ++v) {
        if (g[u][v] == '0') {
          continue;
        }
        if (dist[v] != -1) {
          if (abs(dist[v] - dist[u]) != 1) {
            fail = true;
          }
          continue;
        }
        dist[v] = dist[u] + 1;
        q.push_back(v);
      }
    }
    if (!fail) {
      res = max(res, *max_element(dist.begin(), dist.end()) + 1);
    }
  }
  cout << res << '\n';
}

```