#include<bits/stdc++.h>
using namespace std;
const int N=210;

int gi() {
	int x=0,o=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int n,dep[N];
char s[N];
bool G[N][N];

int bfs(int s) {
	memset(dep,-1,sizeof(dep));
	dep[s]=0;queue<int> q;q.push(s);
	int ans=0;
	while(!q.empty()) {
		int u=q.front();q.pop();
		ans=max(ans,dep[u]+1);
		for(int v=1;v<=n;v++)
			if(G[u][v]) {
				if(!~dep[v]) dep[v]=dep[u]+1,q.push(v);
				else if(abs(dep[u]-dep[v])&1^1) cout<<-1,exit(0);
			}
	}
	return ans;
}

int main() {
	cin>>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';
	}
	int ans=0;
	for(int i=1;i<=n;i++) ans=max(ans,bfs(i));
	cout<<ans;
	return 0;
}

### Prompt

Your challenge is to write 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 N=210;

int gi() {
	int x=0,o=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int n,dep[N];
char s[N];
bool G[N][N];

int bfs(int s) {
	memset(dep,-1,sizeof(dep));
	dep[s]=0;queue<int> q;q.push(s);
	int ans=0;
	while(!q.empty()) {
		int u=q.front();q.pop();
		ans=max(ans,dep[u]+1);
		for(int v=1;v<=n;v++)
			if(G[u][v]) {
				if(!~dep[v]) dep[v]=dep[u]+1,q.push(v);
				else if(abs(dep[u]-dep[v])&1^1) cout<<-1,exit(0);
			}
	}
	return ans;
}

int main() {
	cin>>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';
	}
	int ans=0;
	for(int i=1;i<=n;i++) ans=max(ans,bfs(i));
	cout<<ans;
	return 0;
}

```