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>> adj(N*3, vector<int>{});
	
	for (int i = 0; i < M; ++i)
	{
		int a, b;
		cin >> a >> b;
		a--;
		b--;
		adj[a].push_back(b+N);
		adj[a+N].push_back(b+2*N);
		adj[a+2*N].push_back(b);
	}
	
	int S, T;
	cin >> S >> T;
	S--; T--;
	queue<int> q;
	vector<bool> visited(N*3, false);
	vector<int> distance(N*3, 0);
	
	q.push(S);
	visited[S] = true;
	distance[S] = 0;
	while (!q.empty())
	{
		int s = q.front(); q.pop();
		for (auto x : adj[s])
		{
			if (visited[x])
				continue;
			visited[x] = true;
			distance[x] = distance[s] + 1;
			q.push(x);
		}
	}
	
	if (visited[T])
		cout << distance[T] / 3 << '\n';
	else
		cout << -1 << '\n';
}
