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 <iostream>
#include <vector>
#include <bitset>
#include <queue>
using namespace std;
int main() {
	int N, M;
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> N >> M;
	vector<vector<int> > V(3 * N + 1);
	for(int i = 0; i < M; i++) {
		int u, v;
		cin >> u >> v;
		V[u].push_back(v + N);
		V[u + N].push_back(v + 2 * N);
		V[u + 2 * N].push_back(v);
	}

	vector<int> s(3 * N + 1, -1);

	int S, T;
	cin >> S >> T;
	s[S] = 0;

	queue<int> q;
	q.push(S);
	while(!q.empty()) {
		int p = q.front();
		q.pop();
		for(auto x: V[p]) {
			if(s[x] >= 0) continue;
			s[x] = s[p] + 1;
			if(x == T) {
				cout << s[x] / 3<< endl;
				return 0;
			}
			q.push(x);
		}
	}
	cout << -1 << endl;
}
