#include<iostream>
#include<algorithm>
using lint=int64_t;
using namespace std;

constexpr int INF=1e9;

int N;
string S[210];

int col[210];
bool dfs(int cur=0,int ccol=1)
{
	col[cur]=ccol;
	for(int i=0;i<N;i++)
	{
		if(S[cur][i]=='0')continue;
		if(col[cur]==col[i])
			return false;
		if(col[i]==0 && !dfs(i,-ccol))
			return false;
	}

	return true;
}

int main()
{
	cin >> N;
	for(int i=0;i<N;i++)
		cin >> S[i];

	if(!dfs())
	{
		cout << -1 << endl;
		return 0;
	}

	int dist[210][210]={};
	for(int i=0;i<N;i++)
		for(int j=0;j<N;j++)
			if(i!=j)
				dist[i][j]=INF;
	for(int i=0;i<N;i++)
		for(int j=0;j<N;j++)
			if(S[i][j]=='1')
				dist[i][j]=1;
	
	for(int k=0;k<N;k++)
		for(int i=0;i<N;i++)
			for(int j=0;j<N;j++)
				dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);

	int ans=0;
	for(int i=0;i<N;i++)
		for(int j=0;j<N;j++)
			ans=max(ans,dist[i][j]);
	cout << ans+1 << endl;
	return 0;
}

### Prompt

Create a solution in cpp for 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<iostream>
#include<algorithm>
using lint=int64_t;
using namespace std;

constexpr int INF=1e9;

int N;
string S[210];

int col[210];
bool dfs(int cur=0,int ccol=1)
{
	col[cur]=ccol;
	for(int i=0;i<N;i++)
	{
		if(S[cur][i]=='0')continue;
		if(col[cur]==col[i])
			return false;
		if(col[i]==0 && !dfs(i,-ccol))
			return false;
	}

	return true;
}

int main()
{
	cin >> N;
	for(int i=0;i<N;i++)
		cin >> S[i];

	if(!dfs())
	{
		cout << -1 << endl;
		return 0;
	}

	int dist[210][210]={};
	for(int i=0;i<N;i++)
		for(int j=0;j<N;j++)
			if(i!=j)
				dist[i][j]=INF;
	for(int i=0;i<N;i++)
		for(int j=0;j<N;j++)
			if(S[i][j]=='1')
				dist[i][j]=1;
	
	for(int k=0;k<N;k++)
		for(int i=0;i<N;i++)
			for(int j=0;j<N;j++)
				dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);

	int ans=0;
	for(int i=0;i<N;i++)
		for(int j=0;j<N;j++)
			ans=max(ans,dist[i][j]);
	cout << ans+1 << endl;
	return 0;
}

```