#include<bits/stdc++.h>
using namespace std;

#define MAXN 205
int n,m;
bool a[MAXN][MAXN];
char s[MAXN][MAXN];

void rd()
{
	scanf("%d",&n);
	for(int i = 1; i <= n; i ++) {
		scanf("%s",s[i]+1);
		for(int j = 1; j <= n; j ++)
			a[i][j] = s[i][j]-'0';
	}
}

queue<int>q;
int dis[MAXN];
int ans = -1;

bool bfs(int s)
{
	while(!q.empty()) q.pop();
	memset(dis,-1,sizeof(dis));
	q.push(s);
	dis[s] = 0;
	while(!q.empty())
	{
		int x = q.front();
		q.pop();
		for(int i = 1; i <= n; i ++)
		if(a[x][i]) {
			if(dis[i] == -1) {
				dis[i] = dis[x]+1;
				q.push(i);
			} else {
				if(dis[x] == dis[i]+1 || dis[i] == dis[x]+1) ;
				else return 0;
			}
		}
	}
//	for(int i = 1; i <= n; i ++)
//		cout<<dis[i]<<" ";
//	cout<<"\n";
	for(int i = 1; i <= n; i ++)
		ans = max(ans,dis[i]+1);
	return 1;
}

signed main()
{
	rd();
	for(int i = 1; i <= n; i ++)
		bfs(i);
	cout<<ans;
	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;

#define MAXN 205
int n,m;
bool a[MAXN][MAXN];
char s[MAXN][MAXN];

void rd()
{
	scanf("%d",&n);
	for(int i = 1; i <= n; i ++) {
		scanf("%s",s[i]+1);
		for(int j = 1; j <= n; j ++)
			a[i][j] = s[i][j]-'0';
	}
}

queue<int>q;
int dis[MAXN];
int ans = -1;

bool bfs(int s)
{
	while(!q.empty()) q.pop();
	memset(dis,-1,sizeof(dis));
	q.push(s);
	dis[s] = 0;
	while(!q.empty())
	{
		int x = q.front();
		q.pop();
		for(int i = 1; i <= n; i ++)
		if(a[x][i]) {
			if(dis[i] == -1) {
				dis[i] = dis[x]+1;
				q.push(i);
			} else {
				if(dis[x] == dis[i]+1 || dis[i] == dis[x]+1) ;
				else return 0;
			}
		}
	}
//	for(int i = 1; i <= n; i ++)
//		cout<<dis[i]<<" ";
//	cout<<"\n";
	for(int i = 1; i <= n; i ++)
		ans = max(ans,dis[i]+1);
	return 1;
}

signed main()
{
	rd();
	for(int i = 1; i <= n; i ++)
		bfs(i);
	cout<<ans;
	return 0;
}
```