#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define pb push_back
int n;
vector<int>v[210];
char c[210][210];
int color[210];
bool dfs(int x,int c){
	color[x]=c;
	for(int g:v[x]){
		if(color[g]==c)return false;
		if(color[g]==0&&!dfs(g,-c))return false;
	}
	return true;
}
bool solve(){
	rep(i,n){
		if(color[i]==0){
			if(dfs(i,1)==false)return false;
		}
	}
	return true;
}
int d[210][210];	
int main(){
	cin>>n;
	rep(i,n){
		rep(j,n){
			cin>>c[i][j];
			if(i!=j)d[i][j]=1000000000;
			if(c[i][j]=='1'){
				v[i].pb(j);
				d[i][j]=1;
			}
		}
	}
	if(!solve()){
		puts("-1");
		return 0;
	}
	rep(k,n){
		rep(i,n){
			rep(j,n){
				d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
			}
		}
	}
	int ma=0;
	rep(i,n){
		rep(j,n){
			ma=max(ma,d[i][j]);
		}
	}
	cout<<ma+1<<endl;
}

### Prompt

Generate 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 rep(i,n) for(int i=0;i<n;i++)
#define pb push_back
int n;
vector<int>v[210];
char c[210][210];
int color[210];
bool dfs(int x,int c){
	color[x]=c;
	for(int g:v[x]){
		if(color[g]==c)return false;
		if(color[g]==0&&!dfs(g,-c))return false;
	}
	return true;
}
bool solve(){
	rep(i,n){
		if(color[i]==0){
			if(dfs(i,1)==false)return false;
		}
	}
	return true;
}
int d[210][210];	
int main(){
	cin>>n;
	rep(i,n){
		rep(j,n){
			cin>>c[i][j];
			if(i!=j)d[i][j]=1000000000;
			if(c[i][j]=='1'){
				v[i].pb(j);
				d[i][j]=1;
			}
		}
	}
	if(!solve()){
		puts("-1");
		return 0;
	}
	rep(k,n){
		rep(i,n){
			rep(j,n){
				d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
			}
		}
	}
	int ma=0;
	rep(i,n){
		rep(j,n){
			ma=max(ma,d[i][j]);
		}
	}
	cout<<ma+1<<endl;
}

```