#include<bits/stdc++.h>
using namespace std;
const int maxn = 2012;
char s[maxn];int n;
int g[maxn][maxn],vis[maxn];
void pre_dfs(int x,int c){
	if(vis[x]!=-1){
		if(vis[x]!=c){
			puts("-1");
			exit(0);
		}
		return ;
	}
	vis[x] = c;
	for(int i=1;i<=n;i++)
		if(g[x][i])
			pre_dfs(i,c^1);
}
int main(){
	cin>>n;
	for(int i=1;i<=n;i++){
		scanf("%s",s+1);
		for(int j=1;j<=n;j++){
			if(s[j]=='1')g[i][j] = 1;
			else g[i][j] = 0;
		}
	}
	memset(vis,-1,sizeof(vis));
	pre_dfs(1,0);
	int ans = 0;
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)
		if(g[i][j]==0)g[i][j] = 1024;
	for(int k=1;k<=n;k++)
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)
		g[i][j] = min(g[i][j],g[i][k]+g[k][j]);
	for(int i=1;i<=n;i++)
	for(int j=i+1;j<=n;j++)
	ans = max(ans,g[i][j]);
	cout << ans+1 << endl;
}
### Prompt

Your task is to create 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<bits/stdc++.h>
using namespace std;
const int maxn = 2012;
char s[maxn];int n;
int g[maxn][maxn],vis[maxn];
void pre_dfs(int x,int c){
	if(vis[x]!=-1){
		if(vis[x]!=c){
			puts("-1");
			exit(0);
		}
		return ;
	}
	vis[x] = c;
	for(int i=1;i<=n;i++)
		if(g[x][i])
			pre_dfs(i,c^1);
}
int main(){
	cin>>n;
	for(int i=1;i<=n;i++){
		scanf("%s",s+1);
		for(int j=1;j<=n;j++){
			if(s[j]=='1')g[i][j] = 1;
			else g[i][j] = 0;
		}
	}
	memset(vis,-1,sizeof(vis));
	pre_dfs(1,0);
	int ans = 0;
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)
		if(g[i][j]==0)g[i][j] = 1024;
	for(int k=1;k<=n;k++)
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)
		g[i][j] = min(g[i][j],g[i][k]+g[k][j]);
	for(int i=1;i<=n;i++)
	for(int j=i+1;j<=n;j++)
	ans = max(ans,g[i][j]);
	cout << ans+1 << endl;
}
```