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 (Clang 3.8.0) (judge verdict: Runtime Error)

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

vector< vector<long> > g;
vector< vector<long> > dist;
long n,m,s,t;
long min_d = LONG_MAX;


bool dfs(long i, long d) {
	bool is_t = i==t;
	int j = d%3;

	if ( dist[i][j] >= 0 && dist[i][j]<=d) return is_t;
	dist[i][j] = d;

	if (dist[t][0] != -1) {
		//cout << "min_d : " << min_d << "\t d : " << d << endl;
		min_d = min(dist[t][0], min_d);
		if ( min_d <= d+(3-(j)) ) return is_t;
	}

	//for( auto j : g[i] ) {
	long stop = g[i].size();
	for( long k=0; k<stop; k++) {
		long j = g[i][k];
		bool is_t_tmp = dfs(j, d+1);
		if (!is_t_tmp) {
			if (g[j].size() <= 0) {
				g[i].erase( g[i].begin() + k);
				k--;
				stop--;
			}
		}
		is_t = is_t || is_t_tmp;
	}
	return is_t;
}

int main(void) {
	cin >> n >> m;
	dist = vector< vector<long> > (n+1, vector<long>(3,-1));
	g = vector< vector<long> >(n+1);

	for (long i=0; i<m; i++) {
		long u,v;
		cin >> u >> v;
		g[u].push_back(v);
	}

	cin >> s >> t;

	dfs(s, 0);
	
//	for (long i=1; i<n; i++) {
//		cout << i << " => ";
//		for (auto j : g[i] ) {
//			cout << j << " ";
//		}
//		cout << endl;
//	}


	cout << (dist[t][0]>0 ? dist[t][0]/3 : -1) << endl;
}