#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1e6+10;

vector<int> G[201];
int pos[201], n;

int bfs(int v){
	queue<int> q;
	q.push(v);
	pos[v] = 1;
	int d = 0;
	while(!q.empty()){
		++d;
		int k = q.size();
		while(k--){
			int t = q.front(); q.pop();
			for(int i:G[t]){
				if(pos[i]){
					if(pos[i] != d+1 && pos[i] != d-1) return -1;
				}
				else{
					pos[i] = d+1;
					q.push(i);
				}
			}			
		}
	}
	return d;
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	
	cin >> n;
	char buf[201];
	for(int i=0;i<n;++i){
		cin >> buf;
		for(int j=0;j<n;++j){
			if(buf[j] == '1'){
				G[i].push_back(j);
			}
		}
	}
	int ans = -1;
	for(int i=0;i<n;++i){
		memset(pos, 0, sizeof(pos));
		ans = max(ans, bfs(i));
	}
	cout << ans << endl;
	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>
using namespace std;
using ll = long long;
const int N = 1e6+10;

vector<int> G[201];
int pos[201], n;

int bfs(int v){
	queue<int> q;
	q.push(v);
	pos[v] = 1;
	int d = 0;
	while(!q.empty()){
		++d;
		int k = q.size();
		while(k--){
			int t = q.front(); q.pop();
			for(int i:G[t]){
				if(pos[i]){
					if(pos[i] != d+1 && pos[i] != d-1) return -1;
				}
				else{
					pos[i] = d+1;
					q.push(i);
				}
			}			
		}
	}
	return d;
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	
	cin >> n;
	char buf[201];
	for(int i=0;i<n;++i){
		cin >> buf;
		for(int j=0;j<n;++j){
			if(buf[j] == '1'){
				G[i].push_back(j);
			}
		}
	}
	int ans = -1;
	for(int i=0;i<n;++i){
		memset(pos, 0, sizeof(pos));
		ans = max(ans, bfs(i));
	}
	cout << ans << endl;
	return 0;
}

```