#include <iostream>
#include <queue>
#include <algorithm>
using namespace std;
int n;
char s[200][200];
int d[200][200]; // d[i][j] = (i~jの最短距離)
bool b[200][200];
int ans;

bool bfs(int start){
    queue<int> q;
    q.push(start);
    d[start][start]=0;

    while(q.size()){
        int now=q.front(); q.pop();
        for(int i=0;i<n;i++){
            if(s[now][i]=='0') continue;
            if(d[start][i]==1e8){
                q.push(i);
                d[start][i]=d[start][now]+1;
                ans=max(ans,d[start][i]);
            }else if(d[start][i]==d[start][now]){
                return false;
            }
        }
    }
    return true;
}

int main(){
    cin>>n;
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            cin>>s[i][j];
            d[i][j]=1e8;
        }
    }
    for(int i=0;i<n;i++){
        if(!bfs(i)){
            cout<<-1;
            return 0;
        }
    }
    cout<<ans+1;
}
### 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 <iostream>
#include <queue>
#include <algorithm>
using namespace std;
int n;
char s[200][200];
int d[200][200]; // d[i][j] = (i~jの最短距離)
bool b[200][200];
int ans;

bool bfs(int start){
    queue<int> q;
    q.push(start);
    d[start][start]=0;

    while(q.size()){
        int now=q.front(); q.pop();
        for(int i=0;i<n;i++){
            if(s[now][i]=='0') continue;
            if(d[start][i]==1e8){
                q.push(i);
                d[start][i]=d[start][now]+1;
                ans=max(ans,d[start][i]);
            }else if(d[start][i]==d[start][now]){
                return false;
            }
        }
    }
    return true;
}

int main(){
    cin>>n;
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            cin>>s[i][j];
            d[i][j]=1e8;
        }
    }
    for(int i=0;i<n;i++){
        if(!bfs(i)){
            cout<<-1;
            return 0;
        }
    }
    cout<<ans+1;
}
```