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

int main(){
  ll N; cin >> N;
  vector<vector<ll> > g(N);
  
  for(ll i = 0; i < N; i++){
    for(ll j = 0; j < N; j++){
      char c; cin >> c;
      if(c == '1'){
        g[i].push_back(j);
      }
    }
  }
  
  ll ans = -1;
  
  for(ll i = 0; i < N; i++){
    vector<ll> dist(N, -1);
    queue<ll> que;
    dist[i] = 1;
    que.push(i);
    ll tmp = 0;
    while(!que.empty()){
      ll now = que.front(); que.pop();
      for(auto e: g[now]){
        if(dist[e] == -1){
          dist[e] = dist[now] + 1;
          que.push(e);
        }else if(abs(dist[e] - dist[now]) != 1){
          tmp = -1;
          break;
        }
      }
    }
    if(tmp == -1) continue;
    else{
      for(ll j = 0; j < N; j++){
        tmp = max(tmp, dist[j]);
      }
      ans = max(ans, tmp);
    }
  }
  
  cout << ans << endl;
}
### Prompt

Generate 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;

int main(){
  ll N; cin >> N;
  vector<vector<ll> > g(N);
  
  for(ll i = 0; i < N; i++){
    for(ll j = 0; j < N; j++){
      char c; cin >> c;
      if(c == '1'){
        g[i].push_back(j);
      }
    }
  }
  
  ll ans = -1;
  
  for(ll i = 0; i < N; i++){
    vector<ll> dist(N, -1);
    queue<ll> que;
    dist[i] = 1;
    que.push(i);
    ll tmp = 0;
    while(!que.empty()){
      ll now = que.front(); que.pop();
      for(auto e: g[now]){
        if(dist[e] == -1){
          dist[e] = dist[now] + 1;
          que.push(e);
        }else if(abs(dist[e] - dist[now]) != 1){
          tmp = -1;
          break;
        }
      }
    }
    if(tmp == -1) continue;
    else{
      for(ll j = 0; j < N; j++){
        tmp = max(tmp, dist[j]);
      }
      ans = max(ans, tmp);
    }
  }
  
  cout << ans << endl;
}
```