#include <bits/stdc++.h>
using namespace std;
bool s[205][205];
int v[205];
char c; int n, sol, ans=-1;
deque<int> x;
int main (){
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
cin >> n;
for (int i=1; i<=n; i++){
	for (int j=1; j<=n; j++){
		cin >> c;
		if (c=='1') s[i][j]=1;
	}
}
for (int start=1; start<=n; start++){
	for (int i=1; i<=n; i++) v[i]=0;
	x.push_back(start); sol=0; v[start]=1;
	while (!x.empty()){
		int ob=x.front(); x.pop_front();
		for (int i=1; i<=n; i++){
			if (s[i][ob]){
				if (!v[i]){
					v[i]=v[ob]+1;
					x.push_back(i);
				}
				else if (v[ob]==(v[i]+1) or v[ob]==(v[i]-1)) continue;
				else sol=-1;
			}
		}
		if (sol!=0) break;
	}
	if (sol==0) for (int i=1; i<=n; i++) sol=max(sol, v[i]);
	ans=max(sol, ans);
}
cout << ans;
return 0;
}
### 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>
using namespace std;
bool s[205][205];
int v[205];
char c; int n, sol, ans=-1;
deque<int> x;
int main (){
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
cin >> n;
for (int i=1; i<=n; i++){
	for (int j=1; j<=n; j++){
		cin >> c;
		if (c=='1') s[i][j]=1;
	}
}
for (int start=1; start<=n; start++){
	for (int i=1; i<=n; i++) v[i]=0;
	x.push_back(start); sol=0; v[start]=1;
	while (!x.empty()){
		int ob=x.front(); x.pop_front();
		for (int i=1; i<=n; i++){
			if (s[i][ob]){
				if (!v[i]){
					v[i]=v[ob]+1;
					x.push_back(i);
				}
				else if (v[ob]==(v[i]+1) or v[ob]==(v[i]-1)) continue;
				else sol=-1;
			}
		}
		if (sol!=0) break;
	}
	if (sol==0) for (int i=1; i<=n; i++) sol=max(sol, v[i]);
	ans=max(sol, ans);
}
cout << ans;
return 0;
}
```