#include <bits/stdc++.h>
using namespace std;
#define oo 1000000010
#define mod 1000000007
const int N = 210;
int n;
char grid[N][N];

bool done[N];

int group[N];

int ans = -1;

void start(int node){
	for(int i = 0 ;i < n;i++){
		done[i] = false;
	}
	group[node] = 1;
	queue < int > q;
	done[node] = true;
	q.push(node);
	int mx = 0;
	while(!q.empty()){
		node = q.front();
		q.pop();
		mx = max(mx,group[node]);
		for(int i = 0 ;i < n;i++){
			if(grid[node][i] == '0') continue;
			if(!done[i]){
				done[i] = true;
				group[i] = group[node] + 1;
				q.push(i);
				continue;
			}
			if(abs(group[i] - group[node]) != 1)
				return;
		}
	}
	ans = max(ans , mx);
}

int main(){
	scanf("%d",&n);
	for(int i = 0 ;i < n;i++){
		scanf("%s",grid[i]);
		
	}
	for(int i = 0 ;i < n;i++){
		start(i);
	}
	cout << ans << endl;
	return 0;
}
### Prompt

Your task is to create 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 oo 1000000010
#define mod 1000000007
const int N = 210;
int n;
char grid[N][N];

bool done[N];

int group[N];

int ans = -1;

void start(int node){
	for(int i = 0 ;i < n;i++){
		done[i] = false;
	}
	group[node] = 1;
	queue < int > q;
	done[node] = true;
	q.push(node);
	int mx = 0;
	while(!q.empty()){
		node = q.front();
		q.pop();
		mx = max(mx,group[node]);
		for(int i = 0 ;i < n;i++){
			if(grid[node][i] == '0') continue;
			if(!done[i]){
				done[i] = true;
				group[i] = group[node] + 1;
				q.push(i);
				continue;
			}
			if(abs(group[i] - group[node]) != 1)
				return;
		}
	}
	ans = max(ans , mx);
}

int main(){
	scanf("%d",&n);
	for(int i = 0 ;i < n;i++){
		scanf("%s",grid[i]);
		
	}
	for(int i = 0 ;i < n;i++){
		start(i);
	}
	cout << ans << endl;
	return 0;
}
```