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;

#define rep(i, n) for (int i = 0; i < n; i++)

vector<int> e[300010];

int n;

void add_edge(int from, int to)
{
	int mo=3*n;
	rep(i, 3) e[from+i*n].push_back((to+(i+1)*n)%mo);
}

bool done[300010];
int d[300010];

int main() {
	int m;
	cin >> n >> m;
	rep(i, m)
	{
		int from, to;
		cin >> from >> to;
		from--;
		to--;
		add_edge(from, to);
	}
	int s, t;
	cin >> s >> t;
	s--;
	t--;
	queue<int> q;
	q.push(s);
	d[s]=0;
	while(!q.empty())
	{
		int from=q.front();
		q.pop();
		rep(i, e[from].size())
		{
			int to=e[from][i];
			if(!done[to])
			{
				d[to]=d[from]+1;
				done[to]=true;
				q.push(to);
			}
		}
	}
	if(!done[t]) cout << "-1" << endl;
	else cout << d[t]/3 << endl;
	return 0;
}