#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
ll ans = -2;

int main()
{
ll n ;
cin >> n ;
string s[n];
rep(i,n){
  cin >> s[i];
}
vector<vector<ll>> g(n);
rep(i,n)rep(j,n)if(s[i][j]=='1')g[i].push_back(j);
rep(i,n){
  queue<ll> q;
  q.push(i);
  vector<ll> dist(n,-1);
  dist[i]=0;
  bool ok=1;
  while(!q.empty()){
    ll p=q.front(); q.pop();
    bool okk=1;
    rep(i,g[p].size()){
      if(dist[g[p][i]]==-1){dist[g[p][i]]=dist[p]+1;q.push(g[p][i]);}
      else if(!(dist[g[p][i]]==dist[p]-1 || dist[g[p][i]]==dist[p]+1)){okk=0;break;}
    }
    if(!okk){ok=0;break;}
  }
  if(ok){
    ll te=0;
    rep(i,n)te=max(te,dist[i]);
    ans=max(ans,te);
  }
}


cout << ans+1 << endl;
return 0;
}
### Prompt

Create a solution in Cpp for 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;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
ll ans = -2;

int main()
{
ll n ;
cin >> n ;
string s[n];
rep(i,n){
  cin >> s[i];
}
vector<vector<ll>> g(n);
rep(i,n)rep(j,n)if(s[i][j]=='1')g[i].push_back(j);
rep(i,n){
  queue<ll> q;
  q.push(i);
  vector<ll> dist(n,-1);
  dist[i]=0;
  bool ok=1;
  while(!q.empty()){
    ll p=q.front(); q.pop();
    bool okk=1;
    rep(i,g[p].size()){
      if(dist[g[p][i]]==-1){dist[g[p][i]]=dist[p]+1;q.push(g[p][i]);}
      else if(!(dist[g[p][i]]==dist[p]-1 || dist[g[p][i]]==dist[p]+1)){okk=0;break;}
    }
    if(!okk){ok=0;break;}
  }
  if(ok){
    ll te=0;
    rep(i,n)te=max(te,dist[i]);
    ans=max(ans,te);
  }
}


cout << ans+1 << endl;
return 0;
}
```