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: Wrong Answer)

data class SearchNode(val start: Int, val currentMoveConunt: Int)

fun main(args: Array<String>) {
    val line = readLine()!!.split(' ').map { it.toInt() }
    val nodeNum = line[0]
    val edgeNum = line[1]
    val fromTo = mutableMapOf<Int, MutableList<Int>>()
    repeat(0.until(edgeNum).count()) {
        val l = readLine()!!.split(' ').map { it.toInt() }
        val from = l[0]
        val to = l[1]
        if (fromTo.containsKey(from)) fromTo[from]!!.add(to)
        else fromTo[from] = mutableListOf(to)
    }
    val lastline = readLine()!!.split(' ').map { it.toInt() }
    val start = lastline[0]
    val goal = lastline[1]
    //println("start goal $start   $goal  $fromTo")
    val searchedStart = arrayOfNulls<Boolean>(nodeNum)
    val leastReachCount = arrayOfNulls<Int>(nodeNum)
    for (i in 0.until(nodeNum)) {
        searchedStart[i] = false
        leastReachCount[i] = Int.MAX_VALUE
    }
    val shouldSearchQueue = mutableListOf(SearchNode(start, 0))
    while (shouldSearchQueue.any()) {
        val movestart = shouldSearchQueue.removeAt(0)
        val moveends = move(fromTo, movestart.start)
        //println("moveresult: ${moveends}")
        val nextMoveCount = movestart.currentMoveConunt + 1
        searchedStart[movestart.start - 1] = true
        for (moveend in moveends) {
            if (leastReachCount[moveend - 1]!! > nextMoveCount) {
                leastReachCount[moveend - 1] = nextMoveCount
                if (searchedStart[moveend - 1]!!) continue
                if (shouldSearchQueue.any { it.start == moveend && it.currentMoveConunt <= nextMoveCount }) continue
                shouldSearchQueue.add(0, SearchNode(moveend, nextMoveCount))
            }
        }
    }
    if (leastReachCount[goal - 1] != Int.MAX_VALUE) {
        println(leastReachCount[goal - 1])
    } else {
        println(-1)
    }
}

fun move(fromTo: MutableMap<Int, MutableList<Int>>, from: Int): List<Int> {
    val move1 = fromTo[from] ?: return listOf()
    //println("move1: ${move1}")
    val move2 = move1.map { fromTo[it] }.filter { it != null }.map { it!! }.flatten()
    //println("move2: ${move2}")
    return move2.map { fromTo[it] }.filter { it != null }.map { it!! }.flatten()
}