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: Runtime Error)

package main

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

var mod int = pow(10, 9) + 7
var ans int

func main() {
	reader.Split(bufio.ScanWords)
	N, _ := strconv.Atoi(read())
	M, _ := strconv.Atoi(read())
	to := make([][]int, N+1)
	for i := 0; i < N+1; i++ {
		to[i] = make([]int, 0, N)
	}
	for i := 0; i < M; i++ {
		u, _ := strconv.Atoi(read())
		v, _ := strconv.Atoi(read())
		to[u] = append(to[u], v)
	}
	S, _ := strconv.Atoi(read())
	T, _ := strconv.Atoi(read())

	q := make([][]int, 0, 1e5)
	q = append(q, []int{S, 0})
	var m = [1e5][3]bool{}
	m[S][0] = true
	var step int
	for len(q) != 0 {
		l := len(q)
		nextq := make([][]int, 0, 1e5)
		step++
		for i := 0; i < l; i++ {
			n, w := q[0][0], q[0][1]
			q = q[1:]
			if to[n] != nil {
				for _, value := range to[n] {
					if w == 2 && value == T {
						fmt.Println(step / 3)
						return
					}
					if m[value][(w+1)%3] == false {
						m[value][(w+1)%3] = true
						nextq = append(nextq, []int{value, (w + 1) % 3})
					}
				}
			}
		}
		q = nextq
	}
	fmt.Println(-1)
}

/*  ----------------------------------------  */

var reader = bufio.NewScanner(os.Stdin)

func read() string {
	reader.Scan()
	return reader.Text()
}

func lcm(x, y int) int {
	return (x / gcd(x, y)) * y
}

func gcd(x, y int) int {
	if x%y == 0 {
		return y
	} else {
		r := x % y
		return gcd(y, r)
	}
}

var fac [1000000]int
var finv [1000000]int
var inv [1000000]int

func combination_init() {
	fac[0], fac[1] = 1, 1
	finv[0], finv[1] = 1, 1
	inv[1] = 1

	// invは a^(-1) mod p
	// pをaで割ることを考える

	// p/a*(a) + p%a = p
	// p/a*(a) + p%a = 0          (mod p)
	// -p%a = p/a*(a)             (mod p)
	// -p%a *a^(-1)= p/a          (mod p)
	// a^(-1)= p/a * (-p%a)^(-1)  (mod p)

	for i := 2; i < 1000000; i++ {
		fac[i] = fac[i-1] * i % mod
		inv[i] = mod - inv[mod%i]*(mod/i)%mod
		finv[i] = finv[i-1] * inv[i] % mod
	}
}

func combination(x, y int) int {
	if x < y {
		return 0
	}
	if fac[0] != 1 {
		combination_init()
	}
	return fac[x] * (finv[y] * finv[x-y] % mod) % mod
	//return fac[x] / (fac[y] * fac[x-y])
}

func permutation(x, y int) int {
	if x < y {
		return 0
	}
	if fac[0] != 1 {
		combination_init()
	}
	return fac[x] * (finv[x-y] % mod) % mod
	//return fac[x] / fac[x-y]
}

func max(x ...int) int {
	var res int = x[0]
	for i := 1; i < len(x); i++ {
		res = int(math.Max(float64(x[i]), float64(res)))
	}
	return res
}

func min(x ...int) int {
	var res int = x[0]
	for i := 1; i < len(x); i++ {
		res = int(math.Min(float64(x[i]), float64(res)))
	}
	return res
}
func pow(x, y int) int { return int(math.Pow(float64(x), float64(y))) }
func abs(x int) int    { return int(math.Abs(float64(x))) }
func floor(x int) int  { return int(math.Floor(float64(x))) }
func ceil(x int) int   { return int(math.Ceil(float64(x))) }

type SortBy []int

func (a SortBy) Len() int           { return len(a) }
func (a SortBy) Swap(i, j int)      { a[i], a[j] = a[j], a[i] }
func (a SortBy) Less(i, j int) bool { return a[i] > a[j] }

// 独自の型を定義して、この型に heap interface を実装
type PriorityQueue []int

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