#include <cstdio>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;

const int MaxN = 200;
bool g[MaxN][MaxN]; int n;

int dis[MaxN], ans; queue<int> q;
bool noSolution = false;
void bfs(int from){
	for(int i=0; i<n; ++i)
		dis[i] = 0;
	dis[from] = 1; q.push(from);
	while(not q.empty()){
		int t = q.front(); q.pop();
		ans = max(ans,dis[t]);
		for(int i=0; i<n; ++i){
			if(not g[t][i])
				continue;
			if(not dis[i]){
				dis[i] = dis[t]+1;
				q.push(i);
				continue;
			}
			if(dis[i] < dis[t])
				continue;
			if((dis[i]+dis[t]+1)&1){
				noSolution = true;
				return ;
			}
		}
	}
}

int main(){
	scanf("%d",&n);
	for(int i=0; i<n; ++i){
		getchar();
		for(int j=0; j<n; ++j)
			g[i][j] = (getchar() == '1');
	}
	for(int i=0; i<n; ++i){
		bfs(i);
		if(noSolution){
			puts("-1"); return 0;
		}
	}
	printf("%d",ans);
	return 0;
}
### Prompt

Generate a Cpp solution 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 <cstdio>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;

const int MaxN = 200;
bool g[MaxN][MaxN]; int n;

int dis[MaxN], ans; queue<int> q;
bool noSolution = false;
void bfs(int from){
	for(int i=0; i<n; ++i)
		dis[i] = 0;
	dis[from] = 1; q.push(from);
	while(not q.empty()){
		int t = q.front(); q.pop();
		ans = max(ans,dis[t]);
		for(int i=0; i<n; ++i){
			if(not g[t][i])
				continue;
			if(not dis[i]){
				dis[i] = dis[t]+1;
				q.push(i);
				continue;
			}
			if(dis[i] < dis[t])
				continue;
			if((dis[i]+dis[t]+1)&1){
				noSolution = true;
				return ;
			}
		}
	}
}

int main(){
	scanf("%d",&n);
	for(int i=0; i<n; ++i){
		getchar();
		for(int j=0; j<n; ++j)
			g[i][j] = (getchar() == '1');
	}
	for(int i=0; i<n; ++i){
		bfs(i);
		if(noSolution){
			puts("-1"); return 0;
		}
	}
	printf("%d",ans);
	return 0;
}
```