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 v5.12) (judge verdict: Time Limit Exceeded)

'use strict'

const INF = 900719925474099;
function mapint(temp){
    return temp.split(' ').map((v)=>parseInt(v));
}

function lower_bound(d,a,b,e){
    for(var c;1<b-a;)
        d[c=0|(a+b)/2]<e ? a=c : b=c;
    return d[a]<e?b:a
};

const MOD = 1000000007;

function steps(target,step){


}

function Main(input){
    let temp = input.trim().split('\n');
    let tt = mapint(temp[0]);
    let n = tt[0];
    let m = tt[1];

    let g = [...new Array(n)].map(()=>new Array(0));
    let nt = new Array(n).fill(-1);
    for(let i=1;i<=m;i++){
      tt = mapint(temp[i]);
      g[tt[0]-1].push(tt[1]-1);
    }
    tt = mapint(temp[m+1]);
    let s = tt[0]-1;
    let t = tt[1]-1;
    //console.log(g.map(v=>v.join(' ')),[...nt]);

    let next = [s];
    nt[s] = 0;
    let l = 1;
    let step = 0;
    let f=false;
    let next2 = new Set;
    while(l>0){
    //for(let k=0;k<10;k++){
      next2.clear();
      step++;
      for(let i=0;i<l;i++){
        let ll = g[next[i]].length;
        //console.log('step:',step,' next:',g[next[i]].join(' '));
        for(let j=0;j<ll;j++){
          let gn = g[next[i]][j];
          if(gn == t && step%3 == 0){
            f=true;
            break;
          }
          if(nt[gn] < 0){
            next2.add(gn);
            nt[gn] = step;
          }else{
            if((step - nt[gn])%3 != 0){
              next2.add(gn);
              nt[gn] = step;
            }
          }
        }
        //console.log(nt.join(' '));
        //console.log('---------------------');
        if(f) break;
      }
      if(f) break;
      next = [...next2];
      l = next.length;
      //console.log('loop----------------------------next:',next.join(' '));
    }
    //console.log('exit');
    if(f) console.log(step/3);
    else console.log(-1);
}
//*この行以降は編集しないでください（標準入出力から一度に読み込み、Mainを呼び出します）
Main(require("fs").readFileSync("/dev/stdin", "utf8"));