#include <bits/stdc++.h>
using namespace std;
const int MAXN=210;
int n,flag,ans,vis[MAXN],col[MAXN],s[MAXN][MAXN];
char c[MAXN];
void dfs (int x,int u) {
	col[x]=u;
	for (int i=1;i<=n;i++) {
		if (s[x][i]) {
			if (col[i]==col[x]) {flag=0;}
			else if (!col[i]) {dfs(i,u==1?2:1);}
		}
	}
	return;
}
int bfs (int x) {
	memset(vis,0,sizeof(vis));
	queue <int> q;
	q.push(x);
	int res=0;
	vis[x]=1;
	while (!q.empty()) {
		int a=q.front();
		q.pop();
		res=max(res,vis[a]);
		for (int i=1;i<=n;i++) {
			if (vis[i]) {continue;}
			if (s[a][i]) {vis[i]=vis[a]+1;q.push(i);}
		}
	}
	return res;
}
int main () {
	scanf("%d",&n);
	for (int i=1;i<=n;i++) {
		scanf("%s",c+1);
		for (int j=1;j<=n;j++) {
			s[i][j]=c[j]-'0';
		}
	}
	flag=1;
	dfs(1,1);
	if (!flag) {printf("-1");return 0;}
	for (int i=1;i<=n;i++) {ans=max(ans,bfs(i));}
	printf("%d\n",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 MAXN=210;
int n,flag,ans,vis[MAXN],col[MAXN],s[MAXN][MAXN];
char c[MAXN];
void dfs (int x,int u) {
	col[x]=u;
	for (int i=1;i<=n;i++) {
		if (s[x][i]) {
			if (col[i]==col[x]) {flag=0;}
			else if (!col[i]) {dfs(i,u==1?2:1);}
		}
	}
	return;
}
int bfs (int x) {
	memset(vis,0,sizeof(vis));
	queue <int> q;
	q.push(x);
	int res=0;
	vis[x]=1;
	while (!q.empty()) {
		int a=q.front();
		q.pop();
		res=max(res,vis[a]);
		for (int i=1;i<=n;i++) {
			if (vis[i]) {continue;}
			if (s[a][i]) {vis[i]=vis[a]+1;q.push(i);}
		}
	}
	return res;
}
int main () {
	scanf("%d",&n);
	for (int i=1;i<=n;i++) {
		scanf("%s",c+1);
		for (int j=1;j<=n;j++) {
			s[i][j]=c[j]-'0';
		}
	}
	flag=1;
	dfs(1,1);
	if (!flag) {printf("-1");return 0;}
	for (int i=1;i<=n;i++) {ans=max(ans,bfs(i));}
	printf("%d\n",ans);
	return 0;
}
```