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 (1.14.1) (judge verdict: Accepted)

package main

import "fmt"

func main() {
	var n, m int
	fmt.Scan(&n, &m)

	g := make([][]int, n+1)
	var u, v int
	for i := 0; i < m; i++ {
		fmt.Scan(&u, &v)
		g[u] = append(g[u], v)
	}
	var s, t int
	fmt.Scan(&s, &t)

	ds := make([][3]int, n+1)
	for i := 0; i < n+1; i++ {
		for j := 0; j < 3; j++ {
			ds[i][j] = -1
		}
	}

	q := make([][2]int, 0)
	q = append(q, [2]int{s, 0})
	ds[s][0] = 0
	for len(q) > 0 {
		u := q[0][0]
		d := q[0][1]
		q = q[1:]
		for _, v := range g[u] {
			nd := d + 1
			if ds[v][nd%3] == -1 {
				ds[v][nd%3] = nd
				q = append(q, [2]int{v, nd})
			}
		}
	}

	ans := ds[t][0]
	if ans != -1 {
		ans /= 3
	}
	fmt.Println(ans)
}