#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
const int N=205;

int n;
int dep[N];
char s[N][N];
queue<int>Q;

int Bfs(int st) {
	int ret=1;
	memset(dep,0,sizeof(dep));
	dep[st]=1;Q.push(st);
	while(!Q.empty()) {
		int u=Q.front(); Q.pop();
		ret=max(ret,dep[u]);
		for(int v=1;v<=n;v++)
			if(s[u][v]=='1') {
				if(!dep[v]) {
					dep[v]=dep[u]+1;
					Q.push(v);
				}
				else if(dep[v]!=dep[u]+1&&dep[v]!=dep[u]-1)
					return -1;
			}
	}
	return ret;
}
int main() {
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%s",s[i]+1);
	int ans=-1;
	for(int i=1;i<=n;i++)
		ans=max(ans,Bfs(i));
	printf("%d\n",ans);
	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<cstdio>
#include<queue>
#include<cstring>
using namespace std;
const int N=205;

int n;
int dep[N];
char s[N][N];
queue<int>Q;

int Bfs(int st) {
	int ret=1;
	memset(dep,0,sizeof(dep));
	dep[st]=1;Q.push(st);
	while(!Q.empty()) {
		int u=Q.front(); Q.pop();
		ret=max(ret,dep[u]);
		for(int v=1;v<=n;v++)
			if(s[u][v]=='1') {
				if(!dep[v]) {
					dep[v]=dep[u]+1;
					Q.push(v);
				}
				else if(dep[v]!=dep[u]+1&&dep[v]!=dep[u]-1)
					return -1;
			}
	}
	return ret;
}
int main() {
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%s",s[i]+1);
	int ans=-1;
	for(int i=1;i<=n;i++)
		ans=max(ans,Bfs(i));
	printf("%d\n",ans);
	return 0;
}
```