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: Runtime Error)

const val INF = 1000000000

class Node(private val num: Int) {
    private val adj = mutableListOf<Node>()
    private var dMod0 = INF
    private var dMod1 = INF
    private var dMod2 = INF

    fun adjAdd(v: Node) {
        adj.add(v)
    }

    fun start() {
        dMod0 = 0
    }

    fun dAdd(p: Node): Boolean {
        var result = false
        if (this.dMod0 > p.dMod2 + 1) {
            this.dMod0 = p.dMod2 + 1
            result = true
        }
        if (this.dMod1 > p.dMod0) {
            this.dMod1 = p.dMod0
            result = true
        }
        if (this.dMod2 > p.dMod1) {
            this.dMod2 = p.dMod1
            result = true
        }
        return result
    }

    fun dfs(t: Int): Int {
        var dRet = INF
        if (num == t) dRet = Math.min(dRet, this.dMod0)
        adj.forEach {
            if (it.dAdd(this)) {
                dRet = Math.min(dRet, it.dfs(t))
            }
        }
        return dRet
    }
}

fun main(args: Array<String>){
    val (n, m) = readLine()!!.split(" ").map(String::toInt)
    val graph = mutableListOf(Node(0))
    for (i in 1 until n + 1) graph.add(Node(i))
    for (j in 0 until m) {
        val (u, v) = readLine()!!.split(" ").map(String::toInt)
        graph[u].adjAdd(graph[v])
    }
    val (s, t) = readLine()!!.split(" ").map(String::toInt)
    graph[s].start()
    val ans = graph[s].dfs(t)
    println(if (ans < INF) ans else -1)
}