#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;

const int maxn = 205, inf = int(1e9)+5;

int n, vis[maxn], D[maxn][maxn];
string S[maxn];

bool dfs(int node, int par)
{
	if(vis[node] != -1)
	{
		if(vis[node] != par) return false;
		else return true;
	}
	vis[node] = par;

	for(int i = 0;i < n;i++)
	{
		if(S[node][i] == '1')
		{
			if(!dfs(i, (!par))) return false;
		}
	}

	return true;
}

int main(void)
{
	scanf("%d", &n);

	for(int i = 0;i < n;i++) vis[i] = -1;

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

	if(!dfs(0, 0))
	{
		printf("-1\n");
		return 0;
	}
	
	for(int i = 0;i < n;i++)
	{
		for(int j = 0;j < n;j++) D[i][j] = ((S[i][j] == '0')?inf:1);
		D[i][i] = 0;
	}

	for(int k = 0;k < n;k++)
	{
		for(int i = 0;i < n;i++)
		{
			for(int j = 0;j < n;j++) D[i][j] = min(D[i][k]+D[k][j], D[i][j]);
		}
	}

	int res = 0;
	for(int i = 0;i < n;i++)
	{
		for(int j = 0;j < n;j++) res = max(res, D[i][j]);
	}
	printf("%d\n", res+1);
}
### 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 <iostream>
#include <cstdio>
#include <vector>
using namespace std;

const int maxn = 205, inf = int(1e9)+5;

int n, vis[maxn], D[maxn][maxn];
string S[maxn];

bool dfs(int node, int par)
{
	if(vis[node] != -1)
	{
		if(vis[node] != par) return false;
		else return true;
	}
	vis[node] = par;

	for(int i = 0;i < n;i++)
	{
		if(S[node][i] == '1')
		{
			if(!dfs(i, (!par))) return false;
		}
	}

	return true;
}

int main(void)
{
	scanf("%d", &n);

	for(int i = 0;i < n;i++) vis[i] = -1;

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

	if(!dfs(0, 0))
	{
		printf("-1\n");
		return 0;
	}
	
	for(int i = 0;i < n;i++)
	{
		for(int j = 0;j < n;j++) D[i][j] = ((S[i][j] == '0')?inf:1);
		D[i][i] = 0;
	}

	for(int k = 0;k < n;k++)
	{
		for(int i = 0;i < n;i++)
		{
			for(int j = 0;j < n;j++) D[i][j] = min(D[i][k]+D[k][j], D[i][j]);
		}
	}

	int res = 0;
	for(int i = 0;i < n;i++)
	{
		for(int j = 0;j < n;j++) res = max(res, D[i][j]);
	}
	printf("%d\n", res+1);
}
```