#include <bits/stdc++.h>

using namespace std;

#define sz(x) ((int) ((x).size()))
typedef long long ll;
typedef long double ld;

int n;
vector<int> e[400];

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  cin >> n;
  char c;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      cin >> c;
      if (c == '1')
        e[i].push_back(j);
    }
  int k = -1;
  for (int i = 0; i < n; i++) {
    queue<int> q;
    vector<int> num(n, -1);
    num[i] = 0;
    q.push(i);
    int h = 0;
    bool ok = true;
    while (!q.empty() && ok) {
      int j = q.front();
      q.pop();
      for (int l : e[j]) {
        if (num[l] >= 0)
          ok = ok && abs(num[j] - num[l]) == 1;
        else {
          num[l] = num[j] + 1;
          h = num[l];
          q.push(l);
        }
      }
    }
    if (ok)
      k = max(k, h + 1);
  }
  cout << k << "\n";
}

### 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;

#define sz(x) ((int) ((x).size()))
typedef long long ll;
typedef long double ld;

int n;
vector<int> e[400];

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  cin >> n;
  char c;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      cin >> c;
      if (c == '1')
        e[i].push_back(j);
    }
  int k = -1;
  for (int i = 0; i < n; i++) {
    queue<int> q;
    vector<int> num(n, -1);
    num[i] = 0;
    q.push(i);
    int h = 0;
    bool ok = true;
    while (!q.empty() && ok) {
      int j = q.front();
      q.pop();
      for (int l : e[j]) {
        if (num[l] >= 0)
          ok = ok && abs(num[j] - num[l]) == 1;
        else {
          num[l] = num[j] + 1;
          h = num[l];
          q.push(l);
        }
      }
    }
    if (ok)
      k = max(k, h + 1);
  }
  cout << k << "\n";
}

```