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

const int MAXN = 205;

int N;
char s[MAXN][MAXN];

int dist[MAXN];
int bfs(int src){
	memset(dist, -1, sizeof dist);
	queue <int> q;
	
	q.push(src);
	dist[src] = 0;
	
	int r = 1;
	while(!q.empty()){
		int u = q.front();
		q.pop();
		
		r = max(r, dist[u] + 1);
		
		for(int v = 0; v < N; v++)if(s[u][v] == '1'){
			if(dist[v] == dist[u])return -1;
			if(dist[v] == -1){
				dist[v] = dist[u] + 1;
				q.push(v);
			}
		}
	}
	return r;
}
int main(){	
	scanf("%d", &N);
	for(int i = 0; i < N; i++)
		scanf(" %s", s[i]);
	
	int res = -1;
	for(int i = 0; i < N; i++)
		res = max(res, bfs(i));
	printf("%d\n", res);
	
	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;

const int MAXN = 205;

int N;
char s[MAXN][MAXN];

int dist[MAXN];
int bfs(int src){
	memset(dist, -1, sizeof dist);
	queue <int> q;
	
	q.push(src);
	dist[src] = 0;
	
	int r = 1;
	while(!q.empty()){
		int u = q.front();
		q.pop();
		
		r = max(r, dist[u] + 1);
		
		for(int v = 0; v < N; v++)if(s[u][v] == '1'){
			if(dist[v] == dist[u])return -1;
			if(dist[v] == -1){
				dist[v] = dist[u] + 1;
				q.push(v);
			}
		}
	}
	return r;
}
int main(){	
	scanf("%d", &N);
	for(int i = 0; i < N; i++)
		scanf(" %s", s[i]);
	
	int res = -1;
	for(int i = 0; i < N; i++)
		res = max(res, bfs(i));
	printf("%d\n", res);
	
	return 0;
}

```