#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;
void read(int &x){
    char ch;bool ok;
    for(ok=0,ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-')ok=1;
    for(x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar());if(ok)x=-x;
}
#define rg register
const int maxn=210;queue<int>q;
int n,id[maxn],k,ans,flag;char a[maxn][maxn];
void clear(){while(!q.empty())q.pop();}
void bfs(int now){
    q.push(now);id[now]=1;k=0;
    while(!q.empty()){
	int x=q.front();q.pop();k=max(k,id[x]);
	for(rg int i=1;i<=n;i++)
	    if(a[x][i]=='1'){
		if(id[i]&&id[i]!=id[x]-1&&id[i]!=id[x]+1)return clear(),void();
		if(!id[i])id[i]=id[x]+1,q.push(i);
	    }
    }
    //    printf("%d %d\n",k,now);
    ans=max(ans,k);
}
int main(){
    read(n);
    for(rg int i=1;i<=n;i++)scanf("%s",a[i]+1);
    for(rg int i=1;i<=n;i++)memset(id,0,sizeof id),flag=0,bfs(i);
    printf("%d\n",!ans?-1:ans);
}

### 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<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;
void read(int &x){
    char ch;bool ok;
    for(ok=0,ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-')ok=1;
    for(x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar());if(ok)x=-x;
}
#define rg register
const int maxn=210;queue<int>q;
int n,id[maxn],k,ans,flag;char a[maxn][maxn];
void clear(){while(!q.empty())q.pop();}
void bfs(int now){
    q.push(now);id[now]=1;k=0;
    while(!q.empty()){
	int x=q.front();q.pop();k=max(k,id[x]);
	for(rg int i=1;i<=n;i++)
	    if(a[x][i]=='1'){
		if(id[i]&&id[i]!=id[x]-1&&id[i]!=id[x]+1)return clear(),void();
		if(!id[i])id[i]=id[x]+1,q.push(i);
	    }
    }
    //    printf("%d %d\n",k,now);
    ans=max(ans,k);
}
int main(){
    read(n);
    for(rg int i=1;i<=n;i++)scanf("%s",a[i]+1);
    for(rg int i=1;i<=n;i++)memset(id,0,sizeof id),flag=0,bfs(i);
    printf("%d\n",!ans?-1:ans);
}

```