#include <iostream>
#include <vector>
#include <queue>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int main(){
	int n; cin >> n; vector<int> v[n];
	for(int i=0; i<n; i++) {
		string s; cin >> s;
		for(int j=0; j<n; j++) {
			if (s[j] == '1') {
				v[i].push_back(j);
			}
		}
	}
  int max_an = 0;
  for(int i=0; i<n; i++){
	int done = 1, p[n]; fill(p, p+n, -1); p[i] = 0;
	priority_queue<P, vector<P>, greater<P>> q; q.push({0,i});
	while(!q.empty()) {
		P pp = q.top(); q.pop();
		int fr = pp.second, id = pp.first;
      //cout << id <<" " << fr << endl; 
 		for(int x:v[fr]) {
			if (p[x] == -1) {
				p[x] = id+1; q.push({id+1,x}); done++;
			} else {
				if (p[x] != id+1 && p[x] != id-1) {
				  cout << -1 << endl; return 0;
				}
			}
		}
	}
	int an = 0;
	for(int i=0; i<n; i++) {
		an = max(an, p[i]+1);
	}
    max_an = max(an, max_an);
  }
	cout << max_an << endl;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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 <iostream>
#include <vector>
#include <queue>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int main(){
	int n; cin >> n; vector<int> v[n];
	for(int i=0; i<n; i++) {
		string s; cin >> s;
		for(int j=0; j<n; j++) {
			if (s[j] == '1') {
				v[i].push_back(j);
			}
		}
	}
  int max_an = 0;
  for(int i=0; i<n; i++){
	int done = 1, p[n]; fill(p, p+n, -1); p[i] = 0;
	priority_queue<P, vector<P>, greater<P>> q; q.push({0,i});
	while(!q.empty()) {
		P pp = q.top(); q.pop();
		int fr = pp.second, id = pp.first;
      //cout << id <<" " << fr << endl; 
 		for(int x:v[fr]) {
			if (p[x] == -1) {
				p[x] = id+1; q.push({id+1,x}); done++;
			} else {
				if (p[x] != id+1 && p[x] != id-1) {
				  cout << -1 << endl; return 0;
				}
			}
		}
	}
	int an = 0;
	for(int i=0; i<n; i++) {
		an = max(an, p[i]+1);
	}
    max_an = max(an, max_an);
  }
	cout << max_an << endl;
}
```