#include<bits/stdc++.h>
typedef long long ll;
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
char S[210][210];
int n,G[210][210];
bool col[210],vis[210];
void dfs(int x){
	vis[x]=1;
	for(int i=1;i<=n;++i)
		if(S[x][i]=='1'){
			if(!vis[i])col[i]=!col[x],dfs(i);
			else if(col[i]==col[x])puts("-1"),exit(0);
		}
}
int main(){
#ifdef XZZSB
	freopen("in.in","r",stdin);
	freopen("out.out","w",stdout);
#endif
	n=gi();
	memset(G,63,sizeof G);
	for(int i=1;i<=n;++i)scanf("%s",S[i]+1);
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
			if(S[i][j]=='1'||i==j)G[i][j]=1;
	dfs(1);
	for(int k=1;k<=n;++k)
		for(int i=1;i<=n;++i)
			for(int j=1;j<=n;++j)
				G[i][j]=std::min(G[i][j],G[i][k]+G[k][j]);
	int ans=0;
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
			ans=std::max(ans,G[i][j]);
	printf("%d\n",ans+1);
	return 0;
}

### Prompt

In cpp, your task is to solve 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>
typedef long long ll;
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
char S[210][210];
int n,G[210][210];
bool col[210],vis[210];
void dfs(int x){
	vis[x]=1;
	for(int i=1;i<=n;++i)
		if(S[x][i]=='1'){
			if(!vis[i])col[i]=!col[x],dfs(i);
			else if(col[i]==col[x])puts("-1"),exit(0);
		}
}
int main(){
#ifdef XZZSB
	freopen("in.in","r",stdin);
	freopen("out.out","w",stdout);
#endif
	n=gi();
	memset(G,63,sizeof G);
	for(int i=1;i<=n;++i)scanf("%s",S[i]+1);
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
			if(S[i][j]=='1'||i==j)G[i][j]=1;
	dfs(1);
	for(int k=1;k<=n;++k)
		for(int i=1;i<=n;++i)
			for(int j=1;j<=n;++j)
				G[i][j]=std::min(G[i][j],G[i][k]+G[k][j]);
	int ans=0;
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
			ans=std::max(ans,G[i][j]);
	printf("%d\n",ans+1);
	return 0;
}

```