#include<bits/stdc++.h>
using namespace std;
template<typename T> inline T& Max(T &x,const T &y) {return x<y?(x=y):x;}
inline int read() {
	int x=0,f=1;
	char c=getchar();
	for (;!isdigit(c);c=getchar()) if (c=='-') f=-1;
	for (;isdigit(c);c=getchar()) x=x*10+c-'0';
	return x*f;
}
const int maxn=205;
int n,a[maxn][maxn];
int que[maxn],ql,qr,d[maxn];
int work(int s) {
	que[ql=qr=1]=s;
	fill(d+1,d+n+1,0);
	d[s]=1;
	while (ql<=qr) {
		int x=que[ql++];
		for (int i=1;i<=n;++i) if (a[x][i]) {
			if (!d[i]) {
				d[i]=d[x]+1;
				que[++qr]=i;
				continue;
			}
			if (abs(d[x]-d[i])==1) continue; else return -1;
		}
	}
	return *max_element(d+1,d+n+1);
}
int main() {
	n=read();
	for (int i=1;i<=n;++i) {
		static char s[maxn];
		scanf("%s",s+1);
		for (int j=1;j<=n;++j) a[i][j]=(s[j]=='1');
	}
	int ans(-1);
	for (int i=1;i<=n;++i) {
		int tmp=work(i);
		Max(ans,tmp);
	}
	printf("%d\n",ans);
	return 0;
}

### 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;
template<typename T> inline T& Max(T &x,const T &y) {return x<y?(x=y):x;}
inline int read() {
	int x=0,f=1;
	char c=getchar();
	for (;!isdigit(c);c=getchar()) if (c=='-') f=-1;
	for (;isdigit(c);c=getchar()) x=x*10+c-'0';
	return x*f;
}
const int maxn=205;
int n,a[maxn][maxn];
int que[maxn],ql,qr,d[maxn];
int work(int s) {
	que[ql=qr=1]=s;
	fill(d+1,d+n+1,0);
	d[s]=1;
	while (ql<=qr) {
		int x=que[ql++];
		for (int i=1;i<=n;++i) if (a[x][i]) {
			if (!d[i]) {
				d[i]=d[x]+1;
				que[++qr]=i;
				continue;
			}
			if (abs(d[x]-d[i])==1) continue; else return -1;
		}
	}
	return *max_element(d+1,d+n+1);
}
int main() {
	n=read();
	for (int i=1;i<=n;++i) {
		static char s[maxn];
		scanf("%s",s+1);
		for (int j=1;j<=n;++j) a[i][j]=(s[j]=='1');
	}
	int ans(-1);
	for (int i=1;i<=n;++i) {
		int tmp=work(i);
		Max(ans,tmp);
	}
	printf("%d\n",ans);
	return 0;
}

```