Problem:
Score : 500 points

Problem Statement

Ken loves ken-ken-pa (Japanese version of hopscotch). Today, he will play it on a directed graph G.
G consists of N vertices numbered 1 to N, and M edges. The i-th edge points from Vertex u_i to Vertex v_i.

First, Ken stands on Vertex S. He wants to reach Vertex T by repeating ken-ken-pa. In one ken-ken-pa, he does the following exactly three times: follow an edge pointing from the vertex on which he is standing.

Determine if he can reach Vertex T by repeating ken-ken-pa. If the answer is yes, find the minimum number of ken-ken-pa needed to reach Vertex T. Note that visiting Vertex T in the middle of a ken-ken-pa does not count as reaching Vertex T by repeating ken-ken-pa.

Constraints

2 \leq N \leq 10^5

0 \leq M \leq \min(10^5, N (N-1))

1 \leq u_i, v_i \leq N(1 \leq i \leq M)

u_i \neq v_i (1 \leq i \leq M)

If i \neq j, (u_i, v_i) \neq (u_j, v_j).

1 \leq S, T \leq N

S \neq T

Input

Input is given from Standard Input in the following format:

N M
u_1 v_1
:
u_M v_M
S T

Output

If Ken cannot reach Vertex T from Vertex S by repeating ken-ken-pa, print -1.
If he can, print the minimum number of ken-ken-pa needed to reach vertex T.

Sample Input 1

4 4
1 2
2 3
3 4
4 1
1 3

Sample Output 1

2

Ken can reach Vertex 3 from Vertex 1 in two ken-ken-pa, as follows: 1 \rightarrow 2 \rightarrow 3 \rightarrow 4 in the first ken-ken-pa, then 4 \rightarrow 1 \rightarrow 2 \rightarrow 3 in the second ken-ken-pa. This is the minimum number of ken-ken-pa needed.

Sample Input 2

3 3
1 2
2 3
3 1
1 2

Sample Output 2

-1

Any number of ken-ken-pa will bring Ken back to Vertex 1, so he cannot reach Vertex 2, though he can pass through it in the middle of a ken-ken-pa.

Sample Input 3

2 0
1 2

Sample Output 3

-1

Vertex S and Vertex T may be disconnected.

Sample Input 4

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6

Sample Output 4

2

Solution in JavaScript (Node.js 12.16.1) (judge verdict: Time Limit Exceeded)

//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 one = nextIntArray();
  var N = one[0];
  var M = one[1];
  var list = new Array(N);
  for(var i = 0; i < N; i++){
    list[i] = {
      child : [],
      parent : new Set(),
      access : false,
      depth : 0
    };
  }
  for(var i = 0; i < M; i++){
    var tmp = nextIntArray();
    list[tmp[0] - 1].child.push(tmp[1] - 1);
  }
  for(var i = 0; i < N; i++){
    dfs(i, i, 0);
  }
  var tmp = nextIntArray();
  var s = tmp[0] - 1;
  var t = tmp[1] - 1;
  var queue = [t];
  list[t].access = true;
  var output = 0;
  while(queue.length > 0){
    var now = queue.shift();
    if(now == s){
      output = list[now].depth;
      break;
    }
    var child = Array.from(list[now].parent);
    for(var i = 0; i < child.length; i++){
      if(!list[child[i]].access){
        list[child[i]].access = true;
        list[child[i]].depth = list[now].depth + 1;
        queue.push(child[i]);
      }
    }
  }
  //myerr(list);
  if(output == 0){
    myout(-1);
  }else{
    myout(output);
  }
  

  function dfs(root, now, depth){
    if(depth == 3){
      list[now].parent.add(root);
      return;
    }
    var child = list[now].child;
    for(var j = 0; j < child.length; j++){
      dfs(root, child[j], depth + 1);
    }
    return;
  }
}