#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<ll> vi;
typedef vector<vi> vvi;
#define INF (1e9)

ll bfs(ll v, vvi G, ll N) {
  vi depth = vi(N, -100);
  ll ans = 0;
  depth[v] = 0;
  queue<ll> que;
  que.push(v);
  while (!que.empty()) {
    v = que.front();
    que.pop();
    for (auto next : G[v]) {
      if (depth[next] == depth[v] + 1) continue;
      if (depth[next] == depth[v] - 1) continue;
      if (depth[next] != -100) return -1;
      depth[next] = depth[v] + 1;
      ans = max(ans, depth[next]);
      que.push(next);
    }
  }
  return ans + 1;
}

int main() {
  ll N;
  cin >> N;
  vvi G(N);
  for (ll i = 0; i < N; i++) {
    string s;
    cin >> s;
    for (ll j = 0; j < N; j++) {
      if (s[j] == '0') continue;
      G[i].push_back(j);
      G[j].push_back(i);
    }
  }

  ll ans = -1;
  for (ll i = 0; i < N; i++) {
    ans = max(ans, bfs(i, G, N));
  }
  cout << ans << endl;
}

### 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;
typedef long long ll;
typedef long double ld;
typedef vector<ll> vi;
typedef vector<vi> vvi;
#define INF (1e9)

ll bfs(ll v, vvi G, ll N) {
  vi depth = vi(N, -100);
  ll ans = 0;
  depth[v] = 0;
  queue<ll> que;
  que.push(v);
  while (!que.empty()) {
    v = que.front();
    que.pop();
    for (auto next : G[v]) {
      if (depth[next] == depth[v] + 1) continue;
      if (depth[next] == depth[v] - 1) continue;
      if (depth[next] != -100) return -1;
      depth[next] = depth[v] + 1;
      ans = max(ans, depth[next]);
      que.push(next);
    }
  }
  return ans + 1;
}

int main() {
  ll N;
  cin >> N;
  vvi G(N);
  for (ll i = 0; i < N; i++) {
    string s;
    cin >> s;
    for (ll j = 0; j < N; j++) {
      if (s[j] == '0') continue;
      G[i].push_back(j);
      G[j].push_back(i);
    }
  }

  ll ans = -1;
  for (ll i = 0; i < N; i++) {
    ans = max(ans, bfs(i, G, N));
  }
  cout << ans << endl;
}

```