#include<bits/stdc++.h>
#define db(x) cerr<<#x<<" = "<<x<<endl;
using namespace std;
int sav[205][205];
int dis[205];
int n;
int solve(int s){
	memset(dis,0x3f,sizeof(dis));
	queue<int> q;
	q.push(s);
	int ret=-1;
	dis[s]=1;
	while(!q.empty()){
		int now=q.front();
		q.pop();
		ret=max(ret,dis[now]);
		for(int i=1;i<=n;++i){
			if(sav[now][i]){
				if(dis[i]==0x3f3f3f3f){
					dis[i]=dis[now]+1;
					q.push(i);
				}else if(abs(dis[now]-dis[i])%2==0){
					return -1;
				}
			}
		}
	}
	return ret;
}
int main() {
	cin>>n;
	for(int i=1; i<=n; ++i) {
		for(int j=1; j<=n; ++j) {
			scanf("%1d",&sav[i][j]);
		}
	}
	int ans=-1;
	for(int i=1;i<=n;++i){
		ans=max(ans,solve(i));
	}
	cout<<ans;
	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<bits/stdc++.h>
#define db(x) cerr<<#x<<" = "<<x<<endl;
using namespace std;
int sav[205][205];
int dis[205];
int n;
int solve(int s){
	memset(dis,0x3f,sizeof(dis));
	queue<int> q;
	q.push(s);
	int ret=-1;
	dis[s]=1;
	while(!q.empty()){
		int now=q.front();
		q.pop();
		ret=max(ret,dis[now]);
		for(int i=1;i<=n;++i){
			if(sav[now][i]){
				if(dis[i]==0x3f3f3f3f){
					dis[i]=dis[now]+1;
					q.push(i);
				}else if(abs(dis[now]-dis[i])%2==0){
					return -1;
				}
			}
		}
	}
	return ret;
}
int main() {
	cin>>n;
	for(int i=1; i<=n; ++i) {
		for(int j=1; j<=n; ++j) {
			scanf("%1d",&sav[i][j]);
		}
	}
	int ans=-1;
	for(int i=1;i<=n;++i){
		ans=max(ans,solve(i));
	}
	cout<<ans;
	return 0;
}


```