#include <bits/stdc++.h>

using namespace std;

int n;
const int N=200;
vector<int> v[N];

int ab(int a)
{
    if(a>=0) return a;
    else return -a;
}

int solve(int src)
{
    vector<int> t(n,0);
    t[src]=1;
    queue<int> q;
    q.push(src);
    while(!q.empty())
    {
        int a=q.front();
        q.pop();
        for(int to:v[a])
        {
            if(t[to]==0)
            {
                t[to]=t[a]+1;
                q.push(to);
            }
            else if(ab(t[a]-t[to])!=1) return -1;
        }
    }
    int res=0;
    for(int i=0;i<n;i++) res=max(res,t[i]);
    return res;
}

int main()
{
    scanf("%d",&n);
    char s[n][n+1];
    for(int o=0;o<n;o++) scanf("%s",s[o]);
    for(int o=0;o<n;o++) for(int i=0;i<n;i++) if(s[o][i]=='1') v[o].push_back(i);
    int res=-1;
    for(int i=0;i<n;i++) res=max(res,solve(i));
    printf("%d\n",res);
    return 0;
}

### 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>

using namespace std;

int n;
const int N=200;
vector<int> v[N];

int ab(int a)
{
    if(a>=0) return a;
    else return -a;
}

int solve(int src)
{
    vector<int> t(n,0);
    t[src]=1;
    queue<int> q;
    q.push(src);
    while(!q.empty())
    {
        int a=q.front();
        q.pop();
        for(int to:v[a])
        {
            if(t[to]==0)
            {
                t[to]=t[a]+1;
                q.push(to);
            }
            else if(ab(t[a]-t[to])!=1) return -1;
        }
    }
    int res=0;
    for(int i=0;i<n;i++) res=max(res,t[i]);
    return res;
}

int main()
{
    scanf("%d",&n);
    char s[n][n+1];
    for(int o=0;o<n;o++) scanf("%s",s[o]);
    for(int o=0;o<n;o++) for(int i=0;i<n;i++) if(s[o][i]=='1') v[o].push_back(i);
    int res=-1;
    for(int i=0;i<n;i++) res=max(res,solve(i));
    printf("%d\n",res);
    return 0;
}

```