#define HAVE_STRUCT_TIMESPEC
#include<bits/stdc++.h>
using namespace std;
int col[207],s[207],n;
char ss[207][207];
int bfs(int x){
    int l=0,r=0;
    for(int i=1;i<=n;i++)
        col[i]=0;
    s[++r]=x;
    col[x]=1;
    while(l<r){
        x=s[++l];
        for(int i=1;i<=n;i++)
            if(ss[x][i]-'0'){
                if(col[i]==0){
                    col[i]=col[x]+1;
                    s[++r]=i;
                }
                else if(abs(col[i]-col[x])!=1)
                    return -1;
            }
    }
    return col[s[r]];
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int ans=-1;
    cin>>n;
    for(int i=1;i<=n;++i)
        cin>>ss[i]+1;
    for(int i=1;i<=n;++i)
        ans=max(ans,bfs(i));
    printf("%d",ans);
}

### Prompt

Please provide a cpp coded solution 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
#define HAVE_STRUCT_TIMESPEC
#include<bits/stdc++.h>
using namespace std;
int col[207],s[207],n;
char ss[207][207];
int bfs(int x){
    int l=0,r=0;
    for(int i=1;i<=n;i++)
        col[i]=0;
    s[++r]=x;
    col[x]=1;
    while(l<r){
        x=s[++l];
        for(int i=1;i<=n;i++)
            if(ss[x][i]-'0'){
                if(col[i]==0){
                    col[i]=col[x]+1;
                    s[++r]=i;
                }
                else if(abs(col[i]-col[x])!=1)
                    return -1;
            }
    }
    return col[s[r]];
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int ans=-1;
    cin>>n;
    for(int i=1;i<=n;++i)
        cin>>ss[i]+1;
    for(int i=1;i<=n;++i)
        ans=max(ans,bfs(i));
    printf("%d",ans);
}

```