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 Kotlin (1.0.0) (judge verdict: Time Limit Exceeded)

import java.util.*

fun main(arr:Array<String>) {
    val (n,m) = readLine()!!.split(" ").map { it.toInt() }
    val map = (0 until n).map { MapState(it) }
    for(i in (1..m)) {
        val (a,b) = readLine()!!.split(" ").map { it.toInt() - 1 }
        if(!connect.containsKey(a)) {
            connect[a] = mutableMapOf()
        }
        connect[a]!![b] = true
    }
    val (s,t) = readLine()!!.split(" ").map { it.toInt() - 1 }
    map[s].step[0] = 0
    val queue = LinkedList<Task>()
    queue.push(Task(s, 1, 0))
    while (queue.isNotEmpty()) {
        val current = queue.pop()
        var nextStep = current.step
        var nextSubStep = current.subStep + 1
        if(nextSubStep >= 4) {
            nextStep++
            nextSubStep  = 1
        }
        if(!connect.containsKey(current.id)) {
            continue
        }
        for(next in connect[current.id]!!.keys) {
            if(map[next].step[nextSubStep] <= nextStep) {
                continue
            }
            map[next].step[nextSubStep] = nextStep
            if(next == t && nextSubStep == 3) {
                continue
            }
            queue.push(Task(next, nextStep, nextSubStep))
        }
    }
    val ans = map[t].step[3].let { if(it == Int.MAX_VALUE) -1 else it }
    println(ans)
}
class Task(val id:Int, val step:Int, val subStep:Int)
val connect = mutableMapOf<Int,MutableMap<Int, Boolean>>()

class MapState(val id:Int) {
    val step = (0 until 4).map { Int.MAX_VALUE }.toMutableList()
}