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"
	"container/heap"
	"fmt"
	"math"
	"os"
	"sort"
	"strconv"
)

func getScanner(fp *os.File) *bufio.Scanner {
	scanner := bufio.NewScanner(fp)
	scanner.Split(bufio.ScanWords)
	scanner.Buffer(make([]byte, 500001), 500000)
	return scanner
}

func getNextString(scanner *bufio.Scanner) string {
	scanner.Scan()
	return scanner.Text()
}

func getNextInt(scanner *bufio.Scanner) int {
	i, _ := strconv.Atoi(getNextString(scanner))
	return i
}

func getNextInt64(scanner *bufio.Scanner) int64 {
	i, _ := strconv.ParseInt(getNextString(scanner), 10, 64)
	return i
}

func getNextUint64(scanner *bufio.Scanner) uint64 {
	i, _ := strconv.ParseUint(getNextString(scanner), 10, 64)
	return i
}

func getNextFloat64(scanner *bufio.Scanner) float64 {
	i, _ := strconv.ParseFloat(getNextString(scanner), 64)
	return i
}

func main() {
	fp := os.Stdin
	wfp := os.Stdout

	if os.Getenv("MASPY") == "ますピ" {
		fp, _ = os.Open(os.Getenv("BEET_THE_HARMONY_OF_PERFECT"))
	}
	if os.Getenv("MASPYPY") == "ますピッピ" {
		wfp, _ = os.Create(os.Getenv("NGTKANA_IS_GENIUS10"))
	}

	scanner := getScanner(fp)
	writer := bufio.NewWriter(wfp)

	n := getNextInt(scanner)
	m := getNextInt(scanner)
	g := make([][]int, n)
	gg := make([][]int, n)
	for i := 0; i < n; i++ {
		g[i] = make([]int, 0)
		gg[i] = make([]int, 0)
	}

	for i := 0; i < m; i++ {
		u := getNextInt(scanner) - 1
		v := getNextInt(scanner) - 1
		g[u] = append(g[u], v)
	}

	s := getNextInt(scanner) - 1
	t := getNextInt(scanner) - 1

	var q [4][]int
	for i := 0; i < 4; i++ {
		q[i] = make([]int, 0)
	}
	for i := 0; i < n; i++ {
		for j := 0; j < 4; j++ {
			q[j] = q[j][:0]
		}
		q[0] = append(q[0], i)
		for j := 0; j < 3; j++ {
			for _, from := range q[j] {
				for _, to := range g[from] {
					q[j+1] = append(q[j+1], to)
				}
			}
		}
		sort.Ints(q[3])
		prev := -1
		for _, to := range q[3] {
			if prev == to {
				continue
			}
			if i != to {
				gg[i] = append(gg[i], to)
			}
			prev = to
		}
	}
	nodes := &Nodes{}
	heap.Init(nodes)
	heap.Push(nodes, Node{
		i:    s,
		cost: 0,
	})
	dd := make([]int, n)
	for i := 0; i < n; i++ {
		dd[i] = math.MaxInt32
	}
	visited := make([]int, n)
	for nodes.Len() > 0 {
		p := heap.Pop(nodes).(Node)
		if dd[p.i] < p.cost {
			continue
		}
		dd[p.i] = p.cost
		visited[p.i] = 1
		for _, to := range gg[p.i] {
			if visited[to] == 1 {
				continue
			}
			heap.Push(nodes, Node{
				i:    to,
				cost: p.cost + 1,
			})
		}
	}

	ans := dd[t]
	if ans == math.MaxInt32 {
		ans = -1
	}

	fmt.Fprintln(writer, ans)
	writer.Flush()
}

type Node struct {
	i, cost int
}

type Nodes []Node

func (h Nodes) Len() int           { return len(h) }
func (h Nodes) Less(i, j int) bool { return h[i].cost < h[j].cost }
func (h Nodes) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *Nodes) Push(x interface{}) {
	*h = append(*h, x.(Node))
}
func (h *Nodes) Pop() interface{} {
	old := *h
	n := len(old)
	x := old[n-1]
	*h = old[0 : n-1]
	return x
}