#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e2+5;

int n,k=-1,d[N];
vector<int>g[N];
char s[N][N];

int bfs(int x){
	queue<int>q;
	q.push(x);d[x]=1;
	while(!q.empty()){
		int x=q.front();q.pop();
		for(int i=0;i<g[x].size();++i){
			int y=g[x][i];
			if(d[y]&&abs(d[y]-d[x])!=1)return -1;
			if(!d[y]){
				d[y]=d[x]+1;
				q.push(y);
			}
		}
	}
	return *max_element(d+1,d+n+1);
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)scanf("%s",s[i]+1);
	for(int i=1;i<=n;++i)
	for(int j=1;j<=n;++j)
	if(s[i][j]=='1'){
		g[i].push_back(j);
	}
	for(int i=1;i<=n;++i){
		memset(d,0,sizeof(d));
		k=max(k,bfs(i));
	}
	printf("%d\n",k);
}
### Prompt

In CPP, your task is to solve 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 int N = 2e2+5;

int n,k=-1,d[N];
vector<int>g[N];
char s[N][N];

int bfs(int x){
	queue<int>q;
	q.push(x);d[x]=1;
	while(!q.empty()){
		int x=q.front();q.pop();
		for(int i=0;i<g[x].size();++i){
			int y=g[x][i];
			if(d[y]&&abs(d[y]-d[x])!=1)return -1;
			if(!d[y]){
				d[y]=d[x]+1;
				q.push(y);
			}
		}
	}
	return *max_element(d+1,d+n+1);
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)scanf("%s",s[i]+1);
	for(int i=1;i<=n;++i)
	for(int j=1;j<=n;++j)
	if(s[i][j]=='1'){
		g[i].push_back(j);
	}
	for(int i=1;i<=n;++i){
		memset(d,0,sizeof(d));
		k=max(k,bfs(i));
	}
	printf("%d\n",k);
}
```