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 Haskell (GHC 7.10.3) (judge verdict: Runtime Error)

import Control.Monad
import Data.List
import Data.Array
import Data.Array.ST

main :: IO ()
main = do
  [n,m] <- map (read :: String -> Int) . words <$> getLine
  uvm <- toArray n . map ((\[x,y] -> (x,y)) . map (read :: String -> Int) . words) <$> getNLine_ m
  [s,t] <- map (read :: String -> Int) . words <$> getLine
  let ary = foldl1 (mProduct n) [uvm,uvm,uvm]
  print $ ans n (s,t) ary 1 ary

ans :: Int -> (Int,Int) -> Array (Int,Int) Int -> Int -> Array (Int,Int) Int -> Int
ans n i ary k pary
  | k == n+1 = -1
  | pary ! i /= 0 = k
  | otherwise = ans n i ary (k+1) $ mProduct n pary ary

toArray :: Int -> [(Int,Int)] -> Array (Int,Int) Int
toArray n is = runSTArray $ do
  ary <- newArray ((1,1),(n,n)) 0
  sequence_ [do
    writeArray ary i 1
    | i <- is]
  return ary

mProduct :: Int -> Array (Int,Int) Int -> Array (Int,Int) Int -> Array (Int,Int) Int
mProduct n a b = runSTArray $ do
  ary <- newArray ((1,1),(n,n)) 0
  sequence_ [do
    writeArray ary (s,t) $ func (s,t)
    | s <- [1..n], t <- [1..n]]
  return ary
  where func (s,t) = foldl (\acc k -> (+) acc $ a ! (s,k) * b ! (k,t)) 0 [1..n]

getNLine_ :: Int -> IO [String]
getNLine_ n = replicateM n getLine