#include <cstdio>
#include <cstring>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

const int N=205;

int n,q[N],dep[N];
bool vis[N];
char st[N][N];

int main() {
	scanf("%d",&n);
	fo(i,1,n) scanf("%s",st[i]+1);
	int ans=-1;
	fo(i,1,n) {
		int hd=0,tl=1;q[tl]=i;
		fo(j,1,n) dep[j]=0;dep[i]=1;bool ok=1;
		while (hd<tl) {
			int x=q[++hd];
			fo(j,1,n) 
				if (st[x][j]=='1') {
					if (!dep[j]) dep[j]=dep[x]+1,q[++tl]=j;
					else if (abs(dep[j]-dep[x])!=1) ok=0;
				}
		}
		if (ok) fo(j,1,n) ans=max(ans,dep[j]);
	}
	printf("%d\n",ans);
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 <cstdio>
#include <cstring>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

const int N=205;

int n,q[N],dep[N];
bool vis[N];
char st[N][N];

int main() {
	scanf("%d",&n);
	fo(i,1,n) scanf("%s",st[i]+1);
	int ans=-1;
	fo(i,1,n) {
		int hd=0,tl=1;q[tl]=i;
		fo(j,1,n) dep[j]=0;dep[i]=1;bool ok=1;
		while (hd<tl) {
			int x=q[++hd];
			fo(j,1,n) 
				if (st[x][j]=='1') {
					if (!dep[j]) dep[j]=dep[x]+1,q[++tl]=j;
					else if (abs(dep[j]-dep[x])!=1) ok=0;
				}
		}
		if (ok) fo(j,1,n) ans=max(ans,dep[j]);
	}
	printf("%d\n",ans);
	return 0;
}
```