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.

Examples

Input

4 4
1 2
2 3
3 4
4 1
1 3


Output

2


Input

3 3
1 2
2 3
3 1
1 2


Output

-1


Input

2 0
1 2


Output

-1


Input

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6


Output

2
import sys,os,math
from collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout
from itertools import repeat
import Queue



# n, k = map(int, raw_input().split())
# da = map(int, raw_input().split())
# db = map(int, raw_input().split())


def main():
    n, m = map(int, raw_input().split())
    G = [[] for i in range(n+1)]
    ss = set()
    for i in range(m):
        a, b = map(int, raw_input().split())
        G[a].append(b)
    s, t = map(int, raw_input().split())
    t = t<<2
    q = deque()
    q.append((s<<2, 0))
    ss.add(s<<2)
    while q:
        st, step = q.popleft()
        point, mod = st>>2, st&3
        # print point, step, mod
        if st == t:
            print step/3
            return

        for adj in G[point]:
            new_st = (adj<<2) + (mod+1)%3
            if new_st not in ss:
                ss.add(new_st)
                q.append((new_st, step+1))
    print -1

main()
