#include <bits/stdc++.h>

typedef long long ll;
#define int ll

using namespace std;

const int N=210;

string s[N];
int dist[N];

signed main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int n;
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>s[i];
    int ans=-1;
    for(int ft=0;ft<n;ft++)
    {
        memset(dist,255,sizeof(dist));
        dist[ft]=0;
        queue<int> q;
        q.push(ft);
        while(!q.empty())
        {
            int v=q.front();
            q.pop();
            for(int to=0;to<n;to++)
                if(s[v][to]=='1'&&dist[to]==-1)
                {
                    dist[to]=dist[v]+1;
                    q.push(to);
                }
        }
        bool ok=true;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                if(s[i][j]=='1'&&abs(dist[i]-dist[j])!=1)
                    ok=false;
        if(ok)
            ans=max(ans,*max_element(dist,dist+n)+1);
    }
    cout<<ans;
}

### Prompt

In CPP, your task is to solve 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>

typedef long long ll;
#define int ll

using namespace std;

const int N=210;

string s[N];
int dist[N];

signed main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int n;
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>s[i];
    int ans=-1;
    for(int ft=0;ft<n;ft++)
    {
        memset(dist,255,sizeof(dist));
        dist[ft]=0;
        queue<int> q;
        q.push(ft);
        while(!q.empty())
        {
            int v=q.front();
            q.pop();
            for(int to=0;to<n;to++)
                if(s[v][to]=='1'&&dist[to]==-1)
                {
                    dist[to]=dist[v]+1;
                    q.push(to);
                }
        }
        bool ok=true;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                if(s[i][j]=='1'&&abs(dist[i]-dist[j])!=1)
                    ok=false;
        if(ok)
            ans=max(ans,*max_element(dist,dist+n)+1);
    }
    cout<<ans;
}

```