#include<bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N=200,INF=(1<<30)-1;

int n;
char e[N+9][N+9];

void into(){
  scanf("%d",&n);
  for (int i=1;i<=n;++i)
	scanf("%s",e[i]+1);
}

int dis[N+9][N+9];

void Floyd(){
  for (int i=1;i<=n;++i)
	for (int j=1;j<=n;++j) dis[i][j]=i==j?0:e[i][j]=='0'?INF:1;
  for (int i=1;i<=n;++i)
	for (int j=1;j<=n;++j)
	  for (int k=1;k<=n;++k)
		if (dis[j][i]+dis[i][k]<dis[j][k]) dis[j][k]=dis[j][i]+dis[i][k];
}

int ans;

bool Check_ans(int st){
  for (int i=1;i<=n;++i)
	for (int j=1;j<=n;++j)
	  if (e[i][j]=='1'&&abs(dis[st][i]-dis[st][j])^1) return 0;
  return 1;
}

void Get_ans(){
  ans=-1;
  for (int i=1;i<=n;++i){
	if (!Check_ans(i)) continue;
	for (int j=1;j<=n;++j) ans=max(ans,dis[i][j]+1);
  }
}

void work(){
  Floyd();
  Get_ans();
}

void outo(){
  printf("%d\n",ans);
}

int main(){
  into();
  work();
  outo();
  return 0;
}
### Prompt

Please create a solution in CPP 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;

typedef long long LL;

const int N=200,INF=(1<<30)-1;

int n;
char e[N+9][N+9];

void into(){
  scanf("%d",&n);
  for (int i=1;i<=n;++i)
	scanf("%s",e[i]+1);
}

int dis[N+9][N+9];

void Floyd(){
  for (int i=1;i<=n;++i)
	for (int j=1;j<=n;++j) dis[i][j]=i==j?0:e[i][j]=='0'?INF:1;
  for (int i=1;i<=n;++i)
	for (int j=1;j<=n;++j)
	  for (int k=1;k<=n;++k)
		if (dis[j][i]+dis[i][k]<dis[j][k]) dis[j][k]=dis[j][i]+dis[i][k];
}

int ans;

bool Check_ans(int st){
  for (int i=1;i<=n;++i)
	for (int j=1;j<=n;++j)
	  if (e[i][j]=='1'&&abs(dis[st][i]-dis[st][j])^1) return 0;
  return 1;
}

void Get_ans(){
  ans=-1;
  for (int i=1;i<=n;++i){
	if (!Check_ans(i)) continue;
	for (int j=1;j<=n;++j) ans=max(ans,dis[i][j]+1);
  }
}

void work(){
  Floyd();
  Get_ans();
}

void outo(){
  printf("%d\n",ans);
}

int main(){
  into();
  work();
  outo();
  return 0;
}
```