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: Accepted)

#include <iostream>
#include <vector>
#include <limits.h>
#include <algorithm>
#include <string>
#include <math.h>
#include <limits.h>
#include <queue>
#include <map>
#include <set>
#include <iomanip>
#include <bitset>
#include <cassert>
#include <random>
#include <functional>
#include <stack>
using namespace std;

template<class T>
void in(vector<T>& V) {
	for (auto itr = V.begin(); itr != V.end(); itr++)cin >> *itr;
	return;
}

template<class T>
void out(vector<T>& V) {
	for (T& itr : V)cout << itr << endl;
	return;
}

//重み付きグラフ
//距離の部分をテンプレートに
template<class T>
class wGraph {
public:
	int N;
	vector<vector<pair<int, T>>>* E;
	wGraph(int n, T i) {
		N = n;
		E = new vector<vector<pair<int, T>>>;
		E->resize(N);
		INF = i;
	}
	wGraph() {
	}
	void constructor(int n, T i) {
		N = n;
		E = new vector<vector<pair<int, T>>>;
		E->resize(N);
		INF = i;
	}
	~wGraph() {
		delete E;
	}
	void addEdge(int a, int b, T c) {
		E->at(a).push_back(pair<int, T>(b, c));
	}
	T INF;
};

//ダイクストラする
//条件:重み>0
//グラフ本体、始点、保存用vector
//到達不可能はg.INF
//O(ElogV)
template<class T>
void  dijkstra(wGraph<T>& g, int s, vector<T>& v) {
	v.resize(g.N);
	for (int i = 0; i < g.N; i++) {
		v[i] = g.INF;
	}
	typedef pair<T, int> t;
	priority_queue<t, vector<t>, greater<t>> q;
	q.push(t(0, s));
	while (!q.empty()) {
		t e = q.top();
		q.pop();
		if (v[e.second] != g.INF && v[e.second] <= e.first)continue;
		v[e.second] = e.first;
		for (auto itr = g.E->at(e.second).begin(); itr != g.E->at(e.second).end(); itr++) {
			q.push(t(e.first + (*itr).second, (*itr).first));
		}
	}
	return;
}

int N, M,S,T;
wGraph<int> G;

int main() {
	cin >> N >> M;
	G.constructor(3 * N, INT_MAX);
	for (int m = 0; m < M; m++) {
		int s, t;
		cin >> s >> t;
		s--; t--;
		G.addEdge(s, t + N,1);
		G.addEdge(s + N, t + 2 * N, 1);
		G.addEdge(s + 2 * N, t, 1);
	}
	cin >> S >> T;
	vector<int> ans;
	dijkstra(G, --S, ans);
	if (ans[--T] == INT_MAX)cout << -1 << endl;
	else cout << ans[T]/3 << endl;
	return 0;
}