#include<cstdio>
#include<iostream>
#define RI register int
#define CI const int&
using namespace std;
const int N=205;
int n,q[N],d[N],col[N],ans; bool g[N][N];
inline char get_digit(void)
{
	char ch; while (ch=getchar(),ch!='0'&&ch!='1'); return ch;
}
inline bool BFS(CI st)
{
	RI H=0,T=1,i; for (q[1]=st,i=1;i<=n;++i) col[i]=-1,d[i]=1e9;
	col[st]=0; d[st]=1; while (H<T)
	{
		int now=q[++H]; for (i=1;i<=n;++i) if (g[now][i])
		{
			if (!~col[i]) col[i]=col[now]^1,d[i]=d[now]+1,q[++T]=i;
			else if (col[now]==col[i]) return 0; else d[i]=min(d[i],d[now]+1);
		}
	}
	for (i=1;i<=n;++i) ans=max(ans,d[i]); return 1;
}
int main()
{
	RI i,j; for (scanf("%d",&n),i=1;i<=n;++i)
	for (j=1;j<=n;++j) g[i][j]=get_digit()-48;
	for (i=1;i<=n;++i) if (!BFS(i)) return puts("-1"),0;
	return printf("%d",ans),0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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<iostream>
#define RI register int
#define CI const int&
using namespace std;
const int N=205;
int n,q[N],d[N],col[N],ans; bool g[N][N];
inline char get_digit(void)
{
	char ch; while (ch=getchar(),ch!='0'&&ch!='1'); return ch;
}
inline bool BFS(CI st)
{
	RI H=0,T=1,i; for (q[1]=st,i=1;i<=n;++i) col[i]=-1,d[i]=1e9;
	col[st]=0; d[st]=1; while (H<T)
	{
		int now=q[++H]; for (i=1;i<=n;++i) if (g[now][i])
		{
			if (!~col[i]) col[i]=col[now]^1,d[i]=d[now]+1,q[++T]=i;
			else if (col[now]==col[i]) return 0; else d[i]=min(d[i],d[now]+1);
		}
	}
	for (i=1;i<=n;++i) ans=max(ans,d[i]); return 1;
}
int main()
{
	RI i,j; for (scanf("%d",&n),i=1;i<=n;++i)
	for (j=1;j<=n;++j) g[i][j]=get_digit()-48;
	for (i=1;i<=n;++i) if (!BFS(i)) return puts("-1"),0;
	return printf("%d",ans),0;
}
```