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: Accepted)

fun main(args: Array<String>) {
    val (N, M) = listOfInt()
    val to = Array<MutableList<Int>>(N, { mutableListOf() } )
    repeat(M) {
        val (u, v) = listOfInt(-1)
        to[u].add(v)
    }
    val (S, T) = listOfInt(-1)
    val INF = -1
    var dist = Array<Array<Int>>(N + 1) { Array<Int>(3, { INF }) }

    var q = Queue<Pair<Int, Int>>()
    q.enqueue(Pair(S, 0))
    dist[S][0] = 0
    while (!q.isEmpty()) {
        val e = q.dequeue()!!
        val v = e.first
        val l = e.second
        to[v].forEach {
            val nl = (l + 1) % 3
            if (dist[it][nl] == INF) {
                dist[it][nl] = dist[v][l] + 1
                q.enqueue(Pair(it, nl))
            }
        }
    }
    var ans = dist[T][0]
    if (ans == INF) ans = -1 else ans /= 3
    println(ans)
}

class Queue<T> {
    val elements: MutableList<T> = mutableListOf()
    fun isEmpty() = elements.isEmpty()
    fun count() = elements.size
    fun enqueue(item: T) = elements.add(item)
    fun dequeue() = if (!isEmpty()) elements.removeAt(0) else null
    fun peek() = if (!isEmpty()) elements[0] else null
    override fun toString(): String = elements.toString()
}

fun next() = readLine()!!
fun listOfString() = next().split(" ")
fun listOfInt(delta: Int = 0) = listOfString().map { Integer.parseInt(it) + delta }