#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const ll INF = 1e18;
const int BIG = 1e9 + 555;
const int maxN = 200 + 5;

int n;
char cell[maxN][maxN];

int res = 0;
int layer[maxN];

void bfs(int s){
	fill_n(layer + 1, n, -1);
	queue<int> que;
	que.push(s);
	layer[s] = 1;
	
	while(!que.empty()){
		int u = que.front();
		que.pop();

		for(int v = 1; v <= n; v++){
			if(cell[u][v] == '1'){
				if(layer[v] == -1){
					layer[v] = layer[u] + 1;
					que.push(v);
				} else if(layer[u] == layer[v]){
					cout << "-1\n";
					exit(0);
				}
			}
		}
	}

	res = max(res, *max_element(layer + 1, layer + n + 1));
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	cin >> n;
	for(int i = 1; i <= n; i++){
		cin >> (cell[i] + 1);
	}

	for(int i = 1; i <= n; i++){
		bfs(i);
	}

	cout << res << '\n';

	return 0;
}
### Prompt

Your challenge is to write 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;
const ll INF = 1e18;
const int BIG = 1e9 + 555;
const int maxN = 200 + 5;

int n;
char cell[maxN][maxN];

int res = 0;
int layer[maxN];

void bfs(int s){
	fill_n(layer + 1, n, -1);
	queue<int> que;
	que.push(s);
	layer[s] = 1;
	
	while(!que.empty()){
		int u = que.front();
		que.pop();

		for(int v = 1; v <= n; v++){
			if(cell[u][v] == '1'){
				if(layer[v] == -1){
					layer[v] = layer[u] + 1;
					que.push(v);
				} else if(layer[u] == layer[v]){
					cout << "-1\n";
					exit(0);
				}
			}
		}
	}

	res = max(res, *max_element(layer + 1, layer + n + 1));
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	cin >> n;
	for(int i = 1; i <= n; i++){
		cin >> (cell[i] + 1);
	}

	for(int i = 1; i <= n; i++){
		bfs(i);
	}

	cout << res << '\n';

	return 0;
}
```