#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, N) for (int i = 0; i < (int)N; ++i)
#define all(a) (a).begin(), (a).end()

int N;
vector<vector<int>> G;

int f(int i) {
  vector<int> d(N, N * N);
  d[i] = 1;
  queue<int> q;
  q.push(i);

  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int v : G[u]) {
      if (d[v] < d[u] - 1) return -1;
      if (d[v] == d[u] - 1 || d[v] == d[u] + 1) continue;
      if (d[v] == d[u]) return -1;
      assert(d[v] == N * N);
      d[v] = d[u] + 1;
      q.push(v);
    }
  }
  return *max_element(all(d));
}

int main() {
  cin >> N;
  G.resize(N);
  rep(i, N) rep(j, N) {
    char c;
    cin >> c;
    if (c == '1') {
      G[i].push_back(j);
    };
  }

  int ans = -1;
  rep(i, N) ans = max(ans, f(i));
  cout << ans << endl;

  return 0;
}

### 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;
typedef long long ll;
#define rep(i, N) for (int i = 0; i < (int)N; ++i)
#define all(a) (a).begin(), (a).end()

int N;
vector<vector<int>> G;

int f(int i) {
  vector<int> d(N, N * N);
  d[i] = 1;
  queue<int> q;
  q.push(i);

  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int v : G[u]) {
      if (d[v] < d[u] - 1) return -1;
      if (d[v] == d[u] - 1 || d[v] == d[u] + 1) continue;
      if (d[v] == d[u]) return -1;
      assert(d[v] == N * N);
      d[v] = d[u] + 1;
      q.push(v);
    }
  }
  return *max_element(all(d));
}

int main() {
  cin >> N;
  G.resize(N);
  rep(i, N) rep(j, N) {
    char c;
    cin >> c;
    if (c == '1') {
      G[i].push_back(j);
    };
  }

  int ans = -1;
  rep(i, N) ans = max(ans, f(i));
  cout << ans << endl;

  return 0;
}

```