#include <bits/stdc++.h>
using namespace std;
#define rep(i,s,t) for(int i=s;i<t;i++)
const int inf=1e9+7;
int n,edge[222][222];
void folid()
{
    rep(k,1,n+1)rep(i,1,n+1)rep(j,1,n+1)
    edge[i][j]=min(edge[i][j],edge[i][k]+edge[k][j]);
}
int col[222],flag=1;
void dfs(int now,int co)
{
    if(col[now]&&col[now]!=co){flag=0;return;}
    if(col[now])return;
    col[now]=co;
    rep(i,1,n+1)if(edge[now][i]!=inf&&now!=i)
        dfs(i,co==1?2:1);
}
char s[222];
int main()
{
    scanf("%d",&n);
    rep(i,1,n+1)
    {
        scanf("%s",s+1);
        rep(j,1,n+1)
        if(s[j]=='1')
            edge[i][j]=1;
        else edge[i][j]=inf;
        edge[i][i]=0;
    }
    dfs(1,1);
    if(!flag){printf("-1\n");return 0;}
    folid();
    int res=0;
    rep(i,1,n+1)rep(j,1,n+1)res=max(res,edge[i][j]);
    printf("%d\n",res+1);
}

### Prompt

Your task is to create 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;
#define rep(i,s,t) for(int i=s;i<t;i++)
const int inf=1e9+7;
int n,edge[222][222];
void folid()
{
    rep(k,1,n+1)rep(i,1,n+1)rep(j,1,n+1)
    edge[i][j]=min(edge[i][j],edge[i][k]+edge[k][j]);
}
int col[222],flag=1;
void dfs(int now,int co)
{
    if(col[now]&&col[now]!=co){flag=0;return;}
    if(col[now])return;
    col[now]=co;
    rep(i,1,n+1)if(edge[now][i]!=inf&&now!=i)
        dfs(i,co==1?2:1);
}
char s[222];
int main()
{
    scanf("%d",&n);
    rep(i,1,n+1)
    {
        scanf("%s",s+1);
        rep(j,1,n+1)
        if(s[j]=='1')
            edge[i][j]=1;
        else edge[i][j]=inf;
        edge[i][i]=0;
    }
    dfs(1,1);
    if(!flag){printf("-1\n");return 0;}
    folid();
    int res=0;
    rep(i,1,n+1)rep(j,1,n+1)res=max(res,edge[i][j]);
    printf("%d\n",res+1);
}

```