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 C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

// #define TEST_CASES
using ll = long long;
#define ff first
#define ss second
const int N = 1e5 + 10;

vector<int> G[N], H[N];
int n, m;

bool mem[N][4];
void add_kkp(int u, int o = -1, int s = 3)
{
	if (o == -1) o = u;
	if (mem[u][s])
	{ return; }
	mem[u][s] = 1;
	if (s == 0)
	{
		if (o != u)
			H[o].push_back(u);
		return;
	}
	for (auto v : G[u])
		add_kkp(v, o, s - 1);
}

int bfs(int o, int d)
{
	vector<int> dist(n + 1, 1e9);
	dist[o] = 0;
	queue<int> q;
	q.push(o);
	while (!q.empty())
	{
		int t = q.front();
		q.pop();
		if (t == d) 
			return dist[d];
		for (auto v : H[t])
		{
			if (dist[v] > dist[t] + 1)
				dist[v] = dist[t] + 1,
				q.push(v);
		}
	}
	return -1;
}

void solve()
{
	scanf("%d%d",&n,&m);
	for (int i = 0; i < m; i++)
	{
		int a, b;
		scanf("%d%d",&a,&b);
		G[a].push_back(b);
	}
	int s, t;
	scanf("%d%d",&s,&t);
	for (int i = 1; i <= n; i++)
	{
		memset(mem, 0, sizeof mem);
		add_kkp(i);
	}
	printf("%d\n", bfs(s, t));
}

int main()
{
	int t = 1;
#ifdef TEST_CASES
	cin>>t;
#endif
	while(t--)
	{
		solve();
	}
}