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 Go (1.6) (judge verdict: Time Limit Exceeded)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

type queue []int

func (q queue) Push(p int) queue {
	return append(q, p)
}

func (q queue) Pop() (int, queue) {
	if len(q) == 1 {
		return q[0], make(queue, 0)
	}
	return q[0], q[1:]
}

func main() {
	sc := bufio.NewScanner(os.Stdin)
	sc.Split(bufio.ScanWords)

	sc.Scan()
	N, _ := strconv.Atoi(sc.Text())

	sc.Scan()
	M, _ := strconv.Atoi(sc.Text())

	graph := make([][]int, N)
	for i := 0; i < N; i++ {
		graph[i] = make([]int, 0)
	}

	for i := 0; i < M; i++ {
		sc.Scan()
		ui, _ := strconv.Atoi(sc.Text())

		sc.Scan()
		vi, _ := strconv.Atoi(sc.Text())

		graph[ui-1] = append(graph[ui-1], vi-1)
	}

	twoGraph := make([]map[int]int, N)
	for i := 0; i < N; i++ {
		twoGraph[i] = make(map[int]int)
		for _, t := range graph[i] {
			for _, vertex := range graph[t] {
				twoGraph[i][vertex] = 1
			}
		}
	}

	kenkenGraph := make([]map[int]int, N)
	for i := 0; i < N; i++ {
		kenkenGraph[i] = make(map[int]int)
		for _, t := range graph[i] {
			for vertex, _ := range twoGraph[t] {
				kenkenGraph[i][vertex] = 1
			}
		}
	}

	sc.Scan()
	S, _ := strconv.Atoi(sc.Text())
	S -= 1

	sc.Scan()
	T, _ := strconv.Atoi(sc.Text())
	T -= 1

	dones := make([]int, N)

	que := make(queue, 0)
	que = que.Push(S)
	for len(que) > 0 {
		var current int
		current, que = que.Pop()

		for vertex, _ := range kenkenGraph[current] {
			if vertex != S && dones[vertex] == 0 {
				dones[vertex] += dones[current] + 1
				que = que.Push(vertex)
				if vertex == T {
					goto done
				}
			}
		}
	}

done:
	if dones[T] == 0 {
		fmt.Println(-1)
	} else {
		fmt.Println(dones[T])
	}
}