Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

2
01
10

Sample Output 1

2

We can put Vertex 1 in V_1 and Vertex 2 in V_2.

Sample Input 2

3
011
101
110

Sample Output 2

-1

Sample Input 3

6
010110
101001
010100
101000
100000
010000

Sample Output 3

4

Solution in Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;
use proconio::marker::Chars;
use std::collections::VecDeque;
use std::cmp::max;

fn main()
{
    input! {
    n:usize,
    edge:[Chars;n]
    }

    let mut ans = -1;
    'main: for i in 0..n {
        // 頂点iをv0に配置
        let mut v = vec![-1; n];
        let mut q = VecDeque::new();
        q.push_back((i, 0));
        v[i] = 0;

        let mut mx = 0;
        while let Some((idx, cst)) = q.pop_front() {
            let cst = cst + 1;
            for j in 0..n {
                if edge[idx][j] == '0' {
                    continue;
                }

                if v[j] >= 0 {
                    if cst % 2 != v[j] {
                        // 2分グラフでない
                        continue 'main;
                    }
                    continue;
                }
                v[j] = cst % 2;
                mx = max(mx, cst + 1);
                q.push_back((j, cst));
            }
        }
        ans = max(ans, mx);
    }

    println!("{}", ans);
}