#include <bits/stdc++.h>
using namespace std;
#define pb push_back
vector<vector<int>>adj;
bool che=0;
int col[205];
int vis[205];

void dfs(int v ,int c){
	vis[v]=1;
	col[v]=c;
	for(auto u:adj[v]){
		if(vis[u]==1 && col[u]==col[v]){
			che=1;
		}
		if(vis[u]==0){
			dfs(u, 1-c);
		}
	}
}

int main() {
	int n;
	cin>>n;
	adj.resize(n+1);
	for(int i=1; i<=n; i++){
		string s;
		cin>>s;
		for(int j=0; j<s.size(); j++){
			if(s[j]=='1'){
				adj[i].pb(j+1);
			}
		}
	}
	dfs(1, 1);
	if(che){
		cout<<-1;
		return 0;
	}
	int maxi=0;
	for(int i=1; i<=n; i++){
		int dis[n+1];
		for(int j=1; j<=n; j++){
			dis[j]=1000000000;
		}
		dis[i]=0;
		queue<int>q;
		q.push(i);
		while(!q.empty()){
			int x=q.front();
			q.pop();
			for(auto u:adj[x]){
				if(dis[u]>1+dis[x]){
					dis[u]=1+dis[x];
					q.push(u);
				}
			}
		}
		for(int j=1; j<=n; j++){
			maxi=max(maxi, dis[j]);
		}
	}
	cout<<1+maxi;
	return 0;
}
### 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 <bits/stdc++.h>
using namespace std;
#define pb push_back
vector<vector<int>>adj;
bool che=0;
int col[205];
int vis[205];

void dfs(int v ,int c){
	vis[v]=1;
	col[v]=c;
	for(auto u:adj[v]){
		if(vis[u]==1 && col[u]==col[v]){
			che=1;
		}
		if(vis[u]==0){
			dfs(u, 1-c);
		}
	}
}

int main() {
	int n;
	cin>>n;
	adj.resize(n+1);
	for(int i=1; i<=n; i++){
		string s;
		cin>>s;
		for(int j=0; j<s.size(); j++){
			if(s[j]=='1'){
				adj[i].pb(j+1);
			}
		}
	}
	dfs(1, 1);
	if(che){
		cout<<-1;
		return 0;
	}
	int maxi=0;
	for(int i=1; i<=n; i++){
		int dis[n+1];
		for(int j=1; j<=n; j++){
			dis[j]=1000000000;
		}
		dis[i]=0;
		queue<int>q;
		q.push(i);
		while(!q.empty()){
			int x=q.front();
			q.pop();
			for(auto u:adj[x]){
				if(dis[u]>1+dis[x]){
					dis[u]=1+dis[x];
					q.push(u);
				}
			}
		}
		for(int j=1; j<=n; j++){
			maxi=max(maxi, dis[j]);
		}
	}
	cout<<1+maxi;
	return 0;
}
```