#include<bits/stdc++.h>
using namespace std;
const int N=205;
int n,tot,head[N],nex[N*N*2],to[N*N*2];
void add(int u,int v){to[++tot]=v;nex[tot]=head[u];head[u]=tot;}
char s[N][N];
int vis[N];
int bfs(int s)
{
    memset(vis,0,sizeof(vis));
    queue<int>q;
    q.push(s);
    vis[s]=1;
    int res=0;
    while(!q.empty())
    {
        int u=q.front();q.pop();
        res=max(res,vis[u]);
        for(int i=head[u];i;i=nex[i])
        {
            int v=to[i];
            if(!vis[v]) vis[v]=vis[u]+1,q.push(v);
            else if(abs(vis[v]-vis[u])!=1) return -1;
        }
    }
    return res;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%s",s[i]+1);
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++)
        if(s[i][j]=='1') add(i,j),add(j,i);
    int ans=0;
    for(int i=1;i<=n;i++)
        ans=max(ans,bfs(i));
    if(ans==0) ans=-1;
    printf("%d\n",ans);
}

### Prompt

Please formulate a Cpp solution 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;
const int N=205;
int n,tot,head[N],nex[N*N*2],to[N*N*2];
void add(int u,int v){to[++tot]=v;nex[tot]=head[u];head[u]=tot;}
char s[N][N];
int vis[N];
int bfs(int s)
{
    memset(vis,0,sizeof(vis));
    queue<int>q;
    q.push(s);
    vis[s]=1;
    int res=0;
    while(!q.empty())
    {
        int u=q.front();q.pop();
        res=max(res,vis[u]);
        for(int i=head[u];i;i=nex[i])
        {
            int v=to[i];
            if(!vis[v]) vis[v]=vis[u]+1,q.push(v);
            else if(abs(vis[v]-vis[u])!=1) return -1;
        }
    }
    return res;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%s",s[i]+1);
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++)
        if(s[i][j]=='1') add(i,j),add(j,i);
    int ans=0;
    for(int i=1;i<=n;i++)
        ans=max(ans,bfs(i));
    if(ans==0) ans=-1;
    printf("%d\n",ans);
}

```