#include<bits/stdc++.h>

const int N=205;

int n,a[N][N],dep[N];
char str[N];
std::queue<int> que;

int solve(int x)
{
	memset(dep,0,sizeof(dep));
	while (!que.empty()) que.pop();
	dep[x]=1;que.push(x);
	int ls;
	while (!que.empty())
	{
		int t=que.front();ls=t;que.pop();
		for (int i=1;i<=n;i++)
		{
			if (!a[t][i]) continue;
			if (!dep[i]) dep[i]=dep[t]+1,que.push(i);
			else if (abs(dep[i]-dep[t])!=1) return -1;
		}
	}
	return dep[ls];
}

int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
		scanf("%s",str+1);
		for (int j=1;j<=n;j++) a[i][j]=str[j]-'0';
	}
	int mx=-1;
	for (int i=1;i<=n;i++) mx=std::max(mx,solve(i));
	printf("%d\n",mx);
	return 0;
}
### Prompt

Your challenge is to write 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>

const int N=205;

int n,a[N][N],dep[N];
char str[N];
std::queue<int> que;

int solve(int x)
{
	memset(dep,0,sizeof(dep));
	while (!que.empty()) que.pop();
	dep[x]=1;que.push(x);
	int ls;
	while (!que.empty())
	{
		int t=que.front();ls=t;que.pop();
		for (int i=1;i<=n;i++)
		{
			if (!a[t][i]) continue;
			if (!dep[i]) dep[i]=dep[t]+1,que.push(i);
			else if (abs(dep[i]-dep[t])!=1) return -1;
		}
	}
	return dep[ls];
}

int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
		scanf("%s",str+1);
		for (int j=1;j<=n;j++) a[i][j]=str[j]-'0';
	}
	int mx=-1;
	for (int i=1;i<=n;i++) mx=std::max(mx,solve(i));
	printf("%d\n",mx);
	return 0;
}
```