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;
typedef long long ll;
constexpr int Inf = 1000000001;
constexpr ll INF= 1e18;
constexpr ll MOD = 1000000007;
const double PI = 3.1415926535897;
typedef pair<int,int> P;

struct edge{
	int to,cost;
};

int N,M;
vector<int> graph[100010];
vector<edge> graph2[100010];
int d[100010];

vector<edge> bfs(int a) {
    queue<P> que;
    set<int> st;
    vector<edge> ret;
    que.push(P(a,0));
    while(!que.empty()) {
        P v = que.front();
        que.pop();
        if(v.second == 3) {
            st.insert(v.first);
            continue;
        }
        for(int i = 0;i < graph[v.first].size();i++) {
            que.push(P(graph[v.first].at(i),v.second + 1));
        }
    }
    for(auto x:st) {
        ret.push_back(edge{x,1});
    }
    return ret;
}

void dijkstra(int s) {
	//greater<P>を指定することでfirstが小さい順に取り出せる
	priority_queue<P,vector<P>,greater<P>> que;
	fill(d,d + N,Inf);
	d[s] = 0;
	que.push(P(0,s));

	while(!que.empty()) {
		P p = que.top();
		que.pop();
		int v = p.second;
		if(d[v] < p.first) {
			continue;
		}
		for(int i = 0;i < graph2[v].size();i++) {
			edge e = graph2[v][i];
			if(d[e.to] > d[v] + e.cost) {
				d[e.to] = d[v] + e.cost;
				que.push(P(d[e.to],e.to));
			}
		}
	}
}

int main() {
    cin >> N >> M;
    for(int i = 0;i < M;i++) {
        int A,B;
        cin >> A >> B;
        graph[A - 1].push_back(B - 1);
    }
    for(int i = 0;i < N;i++) {
        graph2[i] = bfs(i);
    }
    int s,t;
    cin >> s >> t;
    s--;
    t--;
    dijkstra(s);
    if(d[t] == Inf) {
        cout << -1 << endl;
    }
    else {
        cout << d[t] << endl;
    }
}