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.

Examples

Input

4 4
1 2
2 3
3 4
4 1
1 3


Output

2


Input

3 3
1 2
2 3
3 1
1 2


Output

-1


Input

2 0
1 2


Output

-1


Input

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6


Output

2
#include <algorithm>
#include <vector>
#include <iostream>
#include <queue>
using namespace std;

int main()
{
	int N, M;
	cin >> N >> M;
	vector<vector<int>> V(3 * N);
	for (int i = 0; i < M; i++) {
		int s, e;
		cin >> s >> e;
		s--, e--;
		V[s].push_back(e + N);
		V[s + N].push_back(e + 2*N);
		V[s + 2 * N].push_back(e);

	}
	int s, t;
	cin >> s >> t;
	s--, t--;
	queue<int> Q;
	vector<int>dist(3 * N,-1);
	dist[s] = 0;
	Q.push(s);
	while (!Q.empty()) {
		int here = Q.front();
		Q.pop();
		for (int i = 0; i < V[here].size(); i++) {
			int next = V[here][i];
			if (dist[next] == -1) {
				dist[next] = dist[here] + 1;
				Q.push(next);
			}
		}
	}
	if (dist[t] == -1)cout << -1;
	else
		cout << dist[t] / 3;
	
	
}