#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 100000 + 10;
const int M = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
const int oo = 1000000000;
#define pb push_back
int n,v[201],an=-1;
char g[201][201];


void go(int u){
	bool vis[201]={};
	v[u]=0;
	queue<int> q;
	q.push(u);
	vis[u]=true;
	int mx=0;
	while(!q.empty()){
		int u=q.front();
		mx=max(mx, v[u]);
		q.pop();
		for(int i=0; i<n; ++i){
			if(g[u][i] && !vis[i]){
				vis[i]=true;
				v[i]=v[u]+1;
				q.push(i);
			}
		}
	}
	bool win=1;
	for(int i=0; i<n; ++i)
		for(int j=0; j<n; ++j)
			if(g[i][j] && abs(v[i]-v[j])!=1)
				win=0;
	if(win)
		an=max(an, mx+1);
}

int main(){
	cin>>n;
	for(int i=0; i<n; ++i){
		scanf("%s", g[i]);
		for(int j=0; j<n; ++j)
			g[i][j]-='0';
	}
	for(int i=0; i<n; ++i)
		go(i);
	cout<<an<<endl;
	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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;
typedef long long ll;
const int N = 100000 + 10;
const int M = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
const int oo = 1000000000;
#define pb push_back
int n,v[201],an=-1;
char g[201][201];


void go(int u){
	bool vis[201]={};
	v[u]=0;
	queue<int> q;
	q.push(u);
	vis[u]=true;
	int mx=0;
	while(!q.empty()){
		int u=q.front();
		mx=max(mx, v[u]);
		q.pop();
		for(int i=0; i<n; ++i){
			if(g[u][i] && !vis[i]){
				vis[i]=true;
				v[i]=v[u]+1;
				q.push(i);
			}
		}
	}
	bool win=1;
	for(int i=0; i<n; ++i)
		for(int j=0; j<n; ++j)
			if(g[i][j] && abs(v[i]-v[j])!=1)
				win=0;
	if(win)
		an=max(an, mx+1);
}

int main(){
	cin>>n;
	for(int i=0; i<n; ++i){
		scanf("%s", g[i]);
		for(int j=0; j<n; ++j)
			g[i][j]-='0';
	}
	for(int i=0; i<n; ++i)
		go(i);
	cout<<an<<endl;
	return 0;
}
```