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)

import java.util.*

fun main(args: Array<String>) {
    data class Node(
            val to: MutableList<Int> = mutableListOf(),
            var visited: MutableMap<Int, Boolean> = mutableMapOf(0 to false, 1 to false, 2 to false),
            var min: MutableMap<Int, Int> = mutableMapOf(0 to Int.MAX_VALUE, 1 to Int.MAX_VALUE, 2 to Int.MAX_VALUE)
    )

    data class Value(
            val index: Int,
            var num: Int,
            var ken: Int
    )
    val (n, m) = readLine()!!.split(' ').map(String::toInt)
    val nodes = mutableMapOf<Int, Node>()
    for (i in 1..n) {
        nodes[i] = Node()
    }
    for (i in 1..m) {
        val (u, v) = readLine()!!.split(' ').map(String::toInt)
        nodes[u]!!.to.add(v)
    }
    val (s, t) = readLine()!!.split(' ').map(String::toInt)
    val queue = ArrayDeque<Value>()
    nodes[s]!!.to.forEach {
        queue.addLast(Value(it, 0, 1))
    }

    while (queue.isNotEmpty()) {
        val value = queue.pollFirst()
        val (index, num, ken) = Triple(value.index, value.num, value.ken)
        val node = nodes[index]!!

        if (node.visited[num]!!) {
            if (node.min[num]!! <= ken) {
                continue
            } else {
                node.min[num] = ken
                if (num == 2) {
                    node.to.forEach { queue.addLast(Value(it, 0, ken + 1)) }
                } else {
                    node.to.forEach { queue.addLast(Value(it, num + 1, ken)) }
                }
            }
        } else {
            node.visited[num] = true
            node.min[num] = ken
            if (num == 2) {
                node.to.forEach { queue.addLast(Value(it, 0, ken + 1)) }
            } else {
                node.to.forEach { queue.addLast(Value(it, num + 1, ken)) }
            }
        }

    }
    if (nodes[t]!!.visited[2]!!) {
        println(nodes[t]!!.min[2])

    } else {
        println(-1)
    }
}