#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n;
char c[201][201];
int d[201][201];
int e[201];
bool dfs(int id){
	for(int i=1; i<=n ;i++){
		if(c[id][i]=='0') continue;
		if(e[id]==e[i]) return false;
		if(e[i]!=0) continue;
		e[i]=3-e[id];
		if(!dfs(i)) return false;
	}
	return true;
}
int main(){
	ios::sync_with_stdio(false);
	cin >> n;
	for(int i=1; i<=n ;i++){
		for(int j=1; j<=n ;j++){
			cin >> c[i][j];
			d[i][j]=1e9;
			if(c[i][j]=='1') d[i][j]=1;
		}
		d[i][i]=0;
	}
	e[1]=1;
	if(!dfs(1)) return cout << "-1\n",0;
	for(int k=1; k<=n ;k++)
	for(int i=1; i<=n ;i++){
		for(int j=1; j<=n ;j++){
			d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
		}
	}
	int ans=0;
	for(int i=1; i<=n ;i++){
		for(int j=1; j<=n ;j++){
			ans=max(ans,d[i][j]);
		}
	}
	cout << ans+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;
typedef long long ll;
int n;
char c[201][201];
int d[201][201];
int e[201];
bool dfs(int id){
	for(int i=1; i<=n ;i++){
		if(c[id][i]=='0') continue;
		if(e[id]==e[i]) return false;
		if(e[i]!=0) continue;
		e[i]=3-e[id];
		if(!dfs(i)) return false;
	}
	return true;
}
int main(){
	ios::sync_with_stdio(false);
	cin >> n;
	for(int i=1; i<=n ;i++){
		for(int j=1; j<=n ;j++){
			cin >> c[i][j];
			d[i][j]=1e9;
			if(c[i][j]=='1') d[i][j]=1;
		}
		d[i][i]=0;
	}
	e[1]=1;
	if(!dfs(1)) return cout << "-1\n",0;
	for(int k=1; k<=n ;k++)
	for(int i=1; i<=n ;i++){
		for(int j=1; j<=n ;j++){
			d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
		}
	}
	int ans=0;
	for(int i=1; i<=n ;i++){
		for(int j=1; j<=n ;j++){
			ans=max(ans,d[i][j]);
		}
	}
	cout << ans+1 << endl;
}
```