#include<bits/stdc++.h>
using namespace std;
#define rep(i,x,y) for(int i=x;i<y;i++)
#define print(A,x,n) rep(I,0,n){cout<<(I ? " ":"")<<A[I]x;}cout<<endl;
#define pprint(A,y,m,n) rep(J,0,m){print(A[J],y,n);}
const long mod=998244353;
const int size=200;
const int inf=1e9;
int N;
vector<int> g[size];
int color[size];
int bfs(int n){
	rep(i,0,N) color[i] = inf;
	color[n] = 1;
	deque<int> dq; dq.push_back(n);
	int k = 0;
	while(!dq.empty()){
		int v = dq.front(); dq.pop_front();
		k = color[v];
		for(int u:g[v]){
			if(color[u] < color[v] - 1 || color[u] == color[v]) return -1;
			if(color[u] == inf){
                color[u] = color[v] + 1;
			    dq.push_back(u);
            }
		}
	}return k;
}
int main(){
	cin>>N;
	string s;
	rep(i,0,N){
		cin>>s;
		rep(j,0,N)if(s[j] == '1') g[i].push_back(j);
	}
	int ans = -1;
	rep(i,0,N) ans = max(ans, bfs(i));
	cout<<ans<<endl;
}
### Prompt

Develop a solution in CPP to the problem described below:

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;
#define rep(i,x,y) for(int i=x;i<y;i++)
#define print(A,x,n) rep(I,0,n){cout<<(I ? " ":"")<<A[I]x;}cout<<endl;
#define pprint(A,y,m,n) rep(J,0,m){print(A[J],y,n);}
const long mod=998244353;
const int size=200;
const int inf=1e9;
int N;
vector<int> g[size];
int color[size];
int bfs(int n){
	rep(i,0,N) color[i] = inf;
	color[n] = 1;
	deque<int> dq; dq.push_back(n);
	int k = 0;
	while(!dq.empty()){
		int v = dq.front(); dq.pop_front();
		k = color[v];
		for(int u:g[v]){
			if(color[u] < color[v] - 1 || color[u] == color[v]) return -1;
			if(color[u] == inf){
                color[u] = color[v] + 1;
			    dq.push_back(u);
            }
		}
	}return k;
}
int main(){
	cin>>N;
	string s;
	rep(i,0,N){
		cin>>s;
		rep(j,0,N)if(s[j] == '1') g[i].push_back(j);
	}
	int ans = -1;
	rep(i,0,N) ans = max(ans, bfs(i));
	cout<<ans<<endl;
}
```