#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;

int m[205][205],p[205];

int floyd(int n){
	for(int k = 1;k <= n;++k){
		for(int i = 1;i <= n;++i){
			for(int j = 1;j <= n;++j){
				m[i][j] = min(m[i][k]+m[k][j],m[i][j]);
			}
		}
	}
	
	int ans = -INF;
	for(int i = 1;i <= n;++i){
		for(int j = 1;j <= n;++j){
			if(i != j) ans = max(m[i][j],ans);
		}
	}
	
	return ans+1;
}

int bfs(int n){
	memset(p,0,sizeof(p));
	queue <int> q;
	p[1] = 1;
	q.push(1);
	
	while(!q.empty()){
		int u = q.front();q.pop();

		for(int i = 1;i <= n;++i){
			if(m[u][i] == 1){
				if(!p[i]) p[i] = p[u]+1,q.push(i);
				else if(abs(p[u] - p[i]) != 1) return 0;
			}
		}
	}
	return 1;
}

int main()
{
	int n,x;
	scanf("%d",&n);
	
	for(int i = 1;i <= n;++i){
		getchar();
		for(int j = 1;j <= n;++j){
			scanf("%1d",&x);
			if(x) m[i][j] = 1;
			else m[i][j] = INF;
		}
	}
	
	if(bfs(n)) printf("%d\n",floyd(n));
	else printf("-1\n");
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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;
const int INF = 0x3f3f3f3f;

int m[205][205],p[205];

int floyd(int n){
	for(int k = 1;k <= n;++k){
		for(int i = 1;i <= n;++i){
			for(int j = 1;j <= n;++j){
				m[i][j] = min(m[i][k]+m[k][j],m[i][j]);
			}
		}
	}
	
	int ans = -INF;
	for(int i = 1;i <= n;++i){
		for(int j = 1;j <= n;++j){
			if(i != j) ans = max(m[i][j],ans);
		}
	}
	
	return ans+1;
}

int bfs(int n){
	memset(p,0,sizeof(p));
	queue <int> q;
	p[1] = 1;
	q.push(1);
	
	while(!q.empty()){
		int u = q.front();q.pop();

		for(int i = 1;i <= n;++i){
			if(m[u][i] == 1){
				if(!p[i]) p[i] = p[u]+1,q.push(i);
				else if(abs(p[u] - p[i]) != 1) return 0;
			}
		}
	}
	return 1;
}

int main()
{
	int n,x;
	scanf("%d",&n);
	
	for(int i = 1;i <= n;++i){
		getchar();
		for(int j = 1;j <= n;++j){
			scanf("%1d",&x);
			if(x) m[i][j] = 1;
			else m[i][j] = INF;
		}
	}
	
	if(bfs(n)) printf("%d\n",floyd(n));
	else printf("-1\n");
	return 0;
}
```