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 LuaJIT (2.0.4) (judge verdict: Accepted)

local n, m = io.read("*n", "*n")
local edge = {}
for i = 1, n do edge[i] = {} end
for i = 1, m do
  local u, v = io.read("*n", "*n")
  table.insert(edge[u], v)
end
local s, t = io.read("*n", "*n")

local len = {}
for i = 1, 3 * n do
  len[i] = -1
end
local start_idx = (s - 1) * 3 + 1
len[start_idx] = 0


local taskstate = {}
for i = 1, 3 * n do taskstate[i] = false end
local tasks = {}
local tasknum = 0
local done = 0
local tasklim = 3 * n

local function addtask(idx)
  if not taskstate[idx] then
    taskstate[idx] = true
    tasknum = tasknum + 1
    local taskidx = tasknum % tasklim
    if taskidx == 0 then taskidx = tasklim end
    tasks[taskidx] = idx
  end
end

local function walk(src, dst)
  if len[dst] < 0 or len[src] + 1 < len[dst] then
    len[dst] = len[src] + 1
    addtask(dst)
  end
end

addtask(start_idx)
local mfl = math.floor

while done < tasknum do
  done = done + 1
  local taskidx = done % tasklim
  if taskidx == 0 then taskidx = tasklim end
  local srcidx = tasks[taskidx]
  taskstate[srcidx] = false
  local pos = 1 + mfl((srcidx - 1) / 3)
  local rem = srcidx - (pos - 1) * 3
  local dstrem = rem + 1
  if dstrem == 4 then dstrem = 1 end
  for i = 1, #edge[pos] do
    local dstpos = edge[pos][i]
    local dstidx = (dstpos - 1) * 3 + dstrem
    walk(srcidx, dstidx)
  end
end
local goal_idx = (t - 1) * 3 + 1
local ret = len[goal_idx]
if 0 < ret then ret = math.floor(ret / 3) end
print(ret)