#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mn=210;
vector<int>g[mn];
int lev[mn];
int main()
{
    cin.tie(0);

    cin.sync_with_stdio(0);
    int n,i,j;
    string s;
    cin>>n;
    for(i=0;i<n;i++){
        cin>>s;
        for(j=0;j<n;j++){
            if(s[j]=='1'){
                g[i].push_back(j);
            }
        }
    }
    int ans=-1;
    for(i=0;i<n;i++){
        bool fail=0;
        memset(lev,0,sizeof(lev));
        queue<int>q;
        q.push(i);
        lev[i]=1;
        while(q.size()){
            int x=q.front();
            q.pop();
            for(int y:g[x]){
                if(lev[y]&&lev[y]!=lev[x]-1&&lev[y]!=lev[x]+1)fail=1;
                else if(!lev[y]){
                    lev[y]=lev[x]+1;
                    q.push(y);
                }
            }
        }
        if(!fail)for(j=0;j<n;j++)ans=max(ans,lev[j]);
    }
    printf("%d",ans);
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const int mn=210;
vector<int>g[mn];
int lev[mn];
int main()
{
    cin.tie(0);

    cin.sync_with_stdio(0);
    int n,i,j;
    string s;
    cin>>n;
    for(i=0;i<n;i++){
        cin>>s;
        for(j=0;j<n;j++){
            if(s[j]=='1'){
                g[i].push_back(j);
            }
        }
    }
    int ans=-1;
    for(i=0;i<n;i++){
        bool fail=0;
        memset(lev,0,sizeof(lev));
        queue<int>q;
        q.push(i);
        lev[i]=1;
        while(q.size()){
            int x=q.front();
            q.pop();
            for(int y:g[x]){
                if(lev[y]&&lev[y]!=lev[x]-1&&lev[y]!=lev[x]+1)fail=1;
                else if(!lev[y]){
                    lev[y]=lev[x]+1;
                    q.push(y);
                }
            }
        }
        if(!fail)for(j=0;j<n;j++)ans=max(ans,lev[j]);
    }
    printf("%d",ans);
}

```