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: Wrong Answer)

package main

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

func wordScanner() *bufio.Scanner {
	sc := bufio.NewScanner(os.Stdin)
	sc.Split(bufio.ScanWords)
	return sc
}

func get_int(sc *bufio.Scanner) int {
	sc.Scan()
	i, err := strconv.Atoi(sc.Text())
	if err != nil {
		panic(err)
	}
	return i
}

var inf int = 123456

func dfs(i, depth int, mins []int, g [][]int) {
	if a, b := depth/3, depth%3; b == 0 {
		if a < mins[i] {
			mins[i] = a
		} else {
			return
		}
	}
	for _, neigh := range g[i] {
		dfs(neigh, depth+1, mins, g)
	}
}

func main() {
	var n, m, s, t int
	sc := wordScanner()
	n = get_int(sc)
	m = get_int(sc)
	us := make([]int, m)
	vs := make([]int, m)
	for i := range us {
		us[i] = get_int(sc) - 1
		vs[i] = get_int(sc) - 1
	}
	s = get_int(sc) - 1
	t = get_int(sc) - 1

	g := make([][]int, n)
	for i := range g {
		g[i] = make([]int, 0)
	}
	for i, u := range us {
		g[u] = append(g[u], vs[i])
	}

	fmt.Println("yahoo!")

	mins := make([]int, n)
	for i := range mins {
		mins[i] = inf
	}
	dfs(s, 0, mins, g)

	if v := mins[t]; v == inf {
		fmt.Println(-1)
	} else {
		fmt.Println(v)
	}
}