#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

int main(){
	int n; cin>>n;
	bool A[200][200];
	rep(u,n){
		string s; cin>>s;
		rep(v,n) A[u][v]=(s[v]=='1');
	}

	int ans=0;
	rep(r,n){
		int d[200];
		rep(u,n) d[u]=-1;
		d[r]=0;
		queue<int> Q;
		Q.push(r);
		while(!Q.empty()){
			int u=Q.front(); Q.pop();
			rep(v,n) if(A[u][v] && d[v]==-1) {
				d[v]=d[u]+1;
				Q.push(v);
			}
		}

		rep(u,n) rep(v,n) if(A[u][v] && abs(d[u]-d[v])!=1) { cout<<-1<<endl; return 0; }
		ans=max(ans,*max_element(d,d+n)+1);
	}
	cout<<ans<<endl;

	return 0;
}

### Prompt

Generate 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>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

int main(){
	int n; cin>>n;
	bool A[200][200];
	rep(u,n){
		string s; cin>>s;
		rep(v,n) A[u][v]=(s[v]=='1');
	}

	int ans=0;
	rep(r,n){
		int d[200];
		rep(u,n) d[u]=-1;
		d[r]=0;
		queue<int> Q;
		Q.push(r);
		while(!Q.empty()){
			int u=Q.front(); Q.pop();
			rep(v,n) if(A[u][v] && d[v]==-1) {
				d[v]=d[u]+1;
				Q.push(v);
			}
		}

		rep(u,n) rep(v,n) if(A[u][v] && abs(d[u]-d[v])!=1) { cout<<-1<<endl; return 0; }
		ans=max(ans,*max_element(d,d+n)+1);
	}
	cout<<ans<<endl;

	return 0;
}

```