#include<bits/stdc++.h>
using namespace std;
#define mp make_pair
const int maxn=200;
char g[maxn+5][maxn+5];
int dis[maxn+5];
int n;
queue<int> q;
int bfs(int S)
{
    while(!q.empty()) q.pop();
    for(int i=1;i<=n;++i) dis[i]=0;
    q.push(S);
    dis[S]=1;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int v=1;v<=n;++v)
        {
            if(g[u][v]=='0') continue;
            if(dis[v]==0)
            {
                dis[v]=dis[u]+1;
                q.push(v);
            }
            else
                if(dis[v]==dis[u]-1) ;
                else
                if(dis[v]!=dis[u]+1) return -1;
        }
    }
    int mx=0;
    for(int i=1;i<=n;++i) mx=max(mx,dis[i]);
    return mx;
}
int solve()
{
    int ans=-1;
    for(int i=1;i<=n;++i) ans=max(ans,bfs(i));
    return ans;
}
int main()
{

    scanf("%d",&n);
    for(int i=1;i<=n;++i)
        scanf("%s",g[i]+1);
    cout<<solve()<<endl;
    return 0;
}


### Prompt

Please create a solution in CPP 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;
#define mp make_pair
const int maxn=200;
char g[maxn+5][maxn+5];
int dis[maxn+5];
int n;
queue<int> q;
int bfs(int S)
{
    while(!q.empty()) q.pop();
    for(int i=1;i<=n;++i) dis[i]=0;
    q.push(S);
    dis[S]=1;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int v=1;v<=n;++v)
        {
            if(g[u][v]=='0') continue;
            if(dis[v]==0)
            {
                dis[v]=dis[u]+1;
                q.push(v);
            }
            else
                if(dis[v]==dis[u]-1) ;
                else
                if(dis[v]!=dis[u]+1) return -1;
        }
    }
    int mx=0;
    for(int i=1;i<=n;++i) mx=max(mx,dis[i]);
    return mx;
}
int solve()
{
    int ans=-1;
    for(int i=1;i<=n;++i) ans=max(ans,bfs(i));
    return ans;
}
int main()
{

    scanf("%d",&n);
    for(int i=1;i<=n;++i)
        scanf("%s",g[i]+1);
    cout<<solve()<<endl;
    return 0;
}


```