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 (1.15.1) (judge verdict: Accepted)

use std::cmp::*;
use std::io::*;

fn dfs(v: usize, c: usize, iro: &mut Vec<usize>, g: &Vec<Vec<usize>>) -> bool {
    iro[v] = c;
    let mut res = true;
    for nv in 0..g.len() {
        if nv != v && g[v][nv] == 1 {
            if iro[nv] == c {
                return false;
            }
            if iro[nv] == 2 {
                res &= dfs(nv, c ^ 1, iro, g);
            }
        }
    }
    res
}

fn main() {
    let mut s: String = String::new();
    stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let mut g: Vec<Vec<usize>> = vec![vec![0; n]; n];
    for i in 0..n {
        let s: Vec<char> = itr.next().unwrap().chars().collect();
        for j in 0..s.len() {
            if s[j] == '0' {
                g[i][j] = 1 << 30;
            } else {
                g[i][j] = 1;
            }
        }
        g[i][i] = 0;
    }
    let mut iro = vec![2; n];
    // cycle check
    if dfs(0, 0, &mut iro, &g) {
        for k in 0..n {
            for i in 0..n {
                for j in 0..n {
                    g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
                }
            }
        }
        let mut ans = 0;

        for i in 0..n {
            for j in 0..n {
                ans = max(ans, g[i][j]);
            }
        }
        println!("{}", ans + 1);
    } else {
        println!("-1");
    }
}