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 <bits/stdc++.h>
using namespace std;




int main(){
	int n, m; cin >> n >> m;
	vector<vector<int>> uv(n);
	
	for (int i = 0; i < m; i++) {
		int u, v; cin >> u >> v;
		uv[u - 1].push_back(v - 1);
	}

	int s, t; cin >> s >> t;

	vector<int> dp[3];
	for(int i = 0; i < 3; i++)
		dp[i].assign(n, -1);
	queue<pair<int, int>> q;
	q.push(make_pair(s - 1, 0));
	while (!q.empty()) {
		pair<int, int> e = q.front();
		q.pop();

		int p = e.second % 3;

		if (dp[p][e.first] != -1)
			continue;

		dp[p][e.first] = e.second;
		for (auto i : uv[e.first]) {
			q.push(make_pair(i, e.second + 1));
		}
	}

	cout <<
		(dp[0][t - 1] == -1 ?
			-1 : dp[0][t - 1] / 3) << endl;

	return 0;
}
