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<cstdio>
#include<vector>
#include<queue>
#define f(x,k) 3*(x-1)+k+1
std::vector<int> a[300005];
int n,m,s,t,dis[300005];
void add(int x,int y){a[x].push_back(y);}
void bfs(int s)
{
	std::queue<int> q;
	for (int i=1;i<=3*n;i++) dis[i]=1000000000;
	dis[s]=0;q.push(s);
	while (!q.empty())
	{
		int v=q.front();q.pop();
		for (int i=0;i<a[v].size();i++)
		{
			int u=a[v][i];
			if (dis[v]+1<dis[u]) dis[u]=dis[v]+1,q.push(u);
	    }
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		add(f(x,0),f(y,1));
		add(f(x,1),f(y,2));
		add(f(x,2),f(y,0));
	}
	scanf("%d%d",&s,&t);
	bfs(f(s,0));
	printf("%d\n",dis[f(t,0)]!=1000000000?dis[f(t,0)]/3:-1);
}