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 <algorithm>
#include <string>
#include <utility>
#include <set>
#include <map>
#include<cmath>
#include <queue>
#include <numeric>

using lli = long long int;
int inf = 1000000007;
using namespace std;

int N, M;
int u, v;
int S, T;

int main()
{
	cin >> N >> M;
	vector<vector<int>> G(3 * N + 1);
	vector<int> cost(3 * N + 1, -3);
	for (int i = 0; i < M; i++) {
		cin >> u >> v;
		G[u].push_back(v + N);
		G[u + N].push_back(v + 2 * N);
		G[u + 2 * N].push_back(v);
	}
	cin >> S >> T;
	queue<int> q;
	q.push(S);
	cost[S] = 0;
	while (q.size() != 0) {
		int now = q.front();
		q.pop();
		for (auto e : G[now]) {
			if (cost[e] == -3) {
				cost[e] = cost[now] + 1;
				q.push(e);
			}
		}
	}
	cout << cost[T] / 3 << endl;
	return 0;
}