#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
#ifdef _debug
#define dout(i) cout << #i << ' ' << i << ' '
#else 
#define dout(i) //
#endif
using ll = long long;
using ull = unsigned long long;
using ul = unsigned;
using db = double;
int n;
const int maxn = 300;
int con[maxn][maxn];
int res = -1;
int pick(int root){
	set<int> last, now, put, next;
	int res = 0;
	now.insert(root);
	for(;now.size();++res){
		for(auto &i : now){
			for(int j = 1;j <= n;++j){
				if(con[i][j]){
					if(!last.count(j)){
						if(put.count(j))return -1;
						next.insert(j);
					}
				}
			}
		}
		for(auto &i : next)put.insert(i);
		last = now;
		now = next;
		next.clear();
	}
	return res;
}
int main(){
	ios_base::sync_with_stdio(0), cin.tie(0);
	cin >> n;
	for(int i = 1;i <= n;++i)for(int j = 1;j <= n;++j){
		char c;
		cin >> c;
		con[i][j] = c == '1';
	}
	for(int i = 1;i <= n;++i)res = max(res, pick(i));
	cout << res << '\n';


}
### Prompt

Please formulate 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>
#pragma GCC optimize("O3")
using namespace std;
#ifdef _debug
#define dout(i) cout << #i << ' ' << i << ' '
#else 
#define dout(i) //
#endif
using ll = long long;
using ull = unsigned long long;
using ul = unsigned;
using db = double;
int n;
const int maxn = 300;
int con[maxn][maxn];
int res = -1;
int pick(int root){
	set<int> last, now, put, next;
	int res = 0;
	now.insert(root);
	for(;now.size();++res){
		for(auto &i : now){
			for(int j = 1;j <= n;++j){
				if(con[i][j]){
					if(!last.count(j)){
						if(put.count(j))return -1;
						next.insert(j);
					}
				}
			}
		}
		for(auto &i : next)put.insert(i);
		last = now;
		now = next;
		next.clear();
	}
	return res;
}
int main(){
	ios_base::sync_with_stdio(0), cin.tie(0);
	cin >> n;
	for(int i = 1;i <= n;++i)for(int j = 1;j <= n;++j){
		char c;
		cin >> c;
		con[i][j] = c == '1';
	}
	for(int i = 1;i <= n;++i)res = max(res, pick(i));
	cout << res << '\n';


}
```