#include <bits/stdc++.h>

int G[205][205],vis[205],n,ans=0,flag=0;
char s[205];
void bfs(int x){
	std::queue<int>q;q.push(x);
	while (!q.empty()){
		int u=q.front();q.pop();
		for(int i=1;i<=n;++i){
			if (!G[u][i])continue;
			if (vis[i]&&vis[u]%2==vis[i]%2){flag=1;return ;}
			if (!vis[i]){vis[i]=vis[u]+1;q.push(i);}
		}
	}
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		scanf("%s",s+1);
		for(int j=1;j<=n;++j)
			G[i][j]=s[j]-'0';
	}
	for(int i=1;i<=n;++i){
		std::memset(vis,0,sizeof(vis));
		vis[i]=1;bfs(i);
		if (flag){printf("-1");return 0;}
		for(int j=1;j<=n;++j)ans=std::max(ans,vis[j]);
	}printf("%d",ans);
	return 0;
}
### Prompt

Create a solution in Cpp for 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>

int G[205][205],vis[205],n,ans=0,flag=0;
char s[205];
void bfs(int x){
	std::queue<int>q;q.push(x);
	while (!q.empty()){
		int u=q.front();q.pop();
		for(int i=1;i<=n;++i){
			if (!G[u][i])continue;
			if (vis[i]&&vis[u]%2==vis[i]%2){flag=1;return ;}
			if (!vis[i]){vis[i]=vis[u]+1;q.push(i);}
		}
	}
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		scanf("%s",s+1);
		for(int j=1;j<=n;++j)
			G[i][j]=s[j]-'0';
	}
	for(int i=1;i<=n;++i){
		std::memset(vis,0,sizeof(vis));
		vis[i]=1;bfs(i);
		if (flag){printf("-1");return 0;}
		for(int j=1;j<=n;++j)ans=std::max(ans,vis[j]);
	}printf("%d",ans);
	return 0;
}
```