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 JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

//Don't have to see. start------------------------------------------
var read = require('readline').createInterface({
  input: process.stdin, output: process.stdout
});
var obj; var inLine = [];
read.on('line', function(input){inLine.push(input);});
read.on('close', function(){
  obj = init(inLine);
  console.error("\n");
  var start = Date.now();
  Main();
  var end = Date.now() - start;
  myerr("time : " + end + "ms");
  myerr("memory : " + Math.round(process.memoryUsage().heapTotal / 1024) + "KB");
});
function makeClone(obj){return JSON.parse(JSON.stringify(obj));}
function nextInt(){return myconv(next(),1);} function nextStrArray(){return myconv(next(),2);}
function nextIntArray(){return myconv(next(),4);} function nextCharArray(){return myconv(next(),6);}
function next(){return obj.next();} function hasNext(){return obj.hasNext();}
function init(input){  
  var returnObj = {
    list : input, index : 0, max : input.length,
    hasNext : function(){return (this.index < this.max);},
    next : function(){if(!this.hasNext()){throw "ArrayIndexOutOfBoundsException ‚There is no more input";}else{var returnInput = this.list[this.index++];return returnInput;}}
  };
  return returnObj;
}
function myout(s){console.log(s);}
function myerr(s){console.error("debug:" + require("util").inspect(s,false,null));}
//param "no" is
//unknown or outlier : return i. 1: parseInt.
//2: split space. 4: split space and parseInt.
//6: split 1 character. 7: split 1 character and parseInt.
//8: join space. 9: join nextline. 0: join no character.
function myconv(i,no){try{switch(no){case 1:return parseInt(i);case 2:return i.split(" ");case 4:return i.split(" ").map(Number);case 6:return i.split("");case 7:return i.split("").map(Number);case 8:return i.join(" ");case 9:return i.join("\n");case 0:return i.join("");default:return i;}}catch(e){return i;}}

//Don't have to see. end------------------------------------------
function Main(){
  var N = nextInt();
  var list = new Array(N);
  for(var i = 0; i < N; i++){
    list[i] = {
      no : i,
      child : [],
      access : false,
      depth : 1
    };
  }
  for(var i = 0; i < N; i++){
    var tmp = nextCharArray();
    for(var j = 0; j < N; j++){
      if(i != j && tmp[j] == "1"){
        list[i].child.push(j);
      }
    }
  }
  
  var output = -1;
  var maxDepth = -1;
  var maeList;
  for(var i = 0; i < N; i++){
    var depth = bfs(i);
    if(maxDepth < depth){
      maeList = list;
      maxDepth = depth;
    }
  }
  
  for(var i = 0; i < N - 1; i++){
    for(var j = i + 1; j < N; j++){
      if(list[i].child.indexOf(j) != -1 && list[j].child.indexOf(i) != -1){
        if(list[i].depth % 2 == list[j].depth % 2){
          myout(-1);
          return;
        }
      }
    }
  }
  myout(maxDepth);
  function bfs(s){
    for(var i = 0; i < N; i++){
      list[i].access = false;
      list[i].depth = 1;
    }
    var queue = [s];
    list[s].access = true;
    var maxDepth = 0;
    while(queue.length > 0){
      var now = queue.shift();
      var child = list[now].child;
      maxDepth = list[now].depth;
      for(var k = 0; k < child.length; k++){
        if(!list[child[k]].access){
          list[child[k]].access = true;
          list[child[k]].depth = list[now].depth + 1;
          queue.push(child[k]);
        }
      }
    }
    return maxDepth;
  }
}