#include<bits/stdc++.h>
using namespace std;
const int N=210;
char s[N][N];
int n,ans;
int col[N],b[N],f[N][N];
void bfs(int now){
	queue<int>q;
	memset(b,0,sizeof(b));
	f[now][now]=0; b[now]=1;
	q.push(now);
	while (!q.empty()) {
		int t=q.front(); q.pop();
		for (int j=1;j<=n;j++)
		 if (!b[j]&&(s[t][j]=='1')) {
		 	f[now][j]=f[now][t]+1;
		 	b[j]=1; q.push(j);
		 }
	}
	for (int i=1;i<=n;i++) ans=max(ans,f[now][i]);
}
bool dfs(int now){
	int p=1;
	for (int i=1;i<=n;i++)
	 if (s[now][i]=='1') {
	 	if (col[i]!=-1&&(col[i]==col[now])) { return 0;}
	 	if (col[i]==-1) {
	 		col[i]=col[now]^1;
	 		p&=dfs(i);
		 }
	 }
	 if (p) return 1; else return 0;
}
bool pan(){
	col[1]=1;
	return dfs(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++) col[i]=-1;
	if (!pan()) {
		puts("-1"); return 0;
	}
	for (int i=1;i<=n;i++) bfs(i);
	printf("%d\n",ans+1);
}
### Prompt

Please create a solution in Cpp 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>
using namespace std;
const int N=210;
char s[N][N];
int n,ans;
int col[N],b[N],f[N][N];
void bfs(int now){
	queue<int>q;
	memset(b,0,sizeof(b));
	f[now][now]=0; b[now]=1;
	q.push(now);
	while (!q.empty()) {
		int t=q.front(); q.pop();
		for (int j=1;j<=n;j++)
		 if (!b[j]&&(s[t][j]=='1')) {
		 	f[now][j]=f[now][t]+1;
		 	b[j]=1; q.push(j);
		 }
	}
	for (int i=1;i<=n;i++) ans=max(ans,f[now][i]);
}
bool dfs(int now){
	int p=1;
	for (int i=1;i<=n;i++)
	 if (s[now][i]=='1') {
	 	if (col[i]!=-1&&(col[i]==col[now])) { return 0;}
	 	if (col[i]==-1) {
	 		col[i]=col[now]^1;
	 		p&=dfs(i);
		 }
	 }
	 if (p) return 1; else return 0;
}
bool pan(){
	col[1]=1;
	return dfs(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++) col[i]=-1;
	if (!pan()) {
		puts("-1"); return 0;
	}
	for (int i=1;i<=n;i++) bfs(i);
	printf("%d\n",ans+1);
}
```