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
#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<cstring>
using namespace std;
const int N=205;
int n,mrk[N],dep[N],ans;vector<int>E[N];queue<int>Q;
void dfs(int u,int c){
	mrk[u]=c;
	for(int v:E[u])
		if(!~mrk[v])dfs(v,c^1);
		else if(mrk[v]!=(c^1))puts("-1"),exit(0);
}
void bfs(int u){
	memset(dep,63,sizeof(dep));
	dep[u]=1;Q.push(u);
	while(!Q.empty()){
		int u=Q.front();Q.pop();
		for(int v:E[u])
			if(dep[v]>dep[u]+1)
				dep[v]=dep[u]+1,Q.push(v);
	}
	for(int i=1;i<=n;++i)ans=max(ans,dep[i]);
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j){
			int x;scanf("%1d",&x);
			if(x)E[i].push_back(j);
		}
	memset(mrk,-1,sizeof(mrk));
	dfs(1,0);
	for(int i=1;i<=n;++i)bfs(i);
	printf("%d\n",ans);
	return 0;
}
