//Created Time:2019年10月05日 星期六 20时41分32秒
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#define N 202

using namespace std;

int n, ans;
int a[N];
char mp[N][N];

int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) scanf("%s", mp[i] + 1);
	for(int i = 1; i <= n; ++i){
		memset(a, 0, sizeof a);
		a[i] = 1; queue < int > Q; Q.push(i);
		while(!Q.empty()){
			int u = Q.front(); Q.pop();
			for(int v = 1; v <= n; ++v){
				if(mp[u][v] == '1'){
					if(a[v] && abs(a[v] - a[u]) != 1){
						puts("-1"); return 0;
					}
					if(!a[v]) a[v] = a[u] + 1, Q.push(v);
				}
			}
		}
		for(int j = 1; j <= n; ++j) ans = max(ans, a[j]);
	}
	cout << ans << endl;
	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
//Created Time:2019年10月05日 星期六 20时41分32秒
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#define N 202

using namespace std;

int n, ans;
int a[N];
char mp[N][N];

int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) scanf("%s", mp[i] + 1);
	for(int i = 1; i <= n; ++i){
		memset(a, 0, sizeof a);
		a[i] = 1; queue < int > Q; Q.push(i);
		while(!Q.empty()){
			int u = Q.front(); Q.pop();
			for(int v = 1; v <= n; ++v){
				if(mp[u][v] == '1'){
					if(a[v] && abs(a[v] - a[u]) != 1){
						puts("-1"); return 0;
					}
					if(!a[v]) a[v] = a[u] + 1, Q.push(v);
				}
			}
		}
		for(int j = 1; j <= n; ++j) ans = max(ans, a[j]);
	}
	cout << ans << endl;
	return 0;
}

```