#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int N=210;
char s[N][N];
int w[N],v[N];
inline int gi() {
    int x=0,o=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') ch=='-'?o=-o:0,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
int main() {
    int n,ans=0;
    cin>>n;
    for(int i=1;i<=n;i++) {
	scanf("%s",s[i]+1);
	for(int j=1;j<=n;j++) s[i][j]-='0';
    }
    for(int i=1;i<=n;i++) {
	int t=0,sum=1;
	memset(w,0,sizeof(w)),w[v[1]=i]=1;
	while(t<sum) {
	    int x=v[++t];
	    for(int y=1;y<=n;y++)
		if(s[x][y]) {
		    if(w[y]) {
			if(~(w[x]^w[y])&1) return cout<<-1,0;
		    }
		    else w[y]=w[x]+1,v[++sum]=y;
		}
	}
	ans=max(ans,w[v[sum]]);
    }
    cout<<ans;
    return 0;
}
### Prompt

Generate 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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int N=210;
char s[N][N];
int w[N],v[N];
inline int gi() {
    int x=0,o=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') ch=='-'?o=-o:0,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
int main() {
    int n,ans=0;
    cin>>n;
    for(int i=1;i<=n;i++) {
	scanf("%s",s[i]+1);
	for(int j=1;j<=n;j++) s[i][j]-='0';
    }
    for(int i=1;i<=n;i++) {
	int t=0,sum=1;
	memset(w,0,sizeof(w)),w[v[1]=i]=1;
	while(t<sum) {
	    int x=v[++t];
	    for(int y=1;y<=n;y++)
		if(s[x][y]) {
		    if(w[y]) {
			if(~(w[x]^w[y])&1) return cout<<-1,0;
		    }
		    else w[y]=w[x]+1,v[++sum]=y;
		}
	}
	ans=max(ans,w[v[sum]]);
    }
    cout<<ans;
    return 0;
}
```