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>
#define inf ((int)(1e9))
using namespace std;
int n,m,s,t;
vector<int> graph[100005];
int main()
{
	cin>>n>>m;

	for(int i=0;i<m;i++)
	{
		int u,v;
		cin>>u>>v;
		--u,--v;
		graph[u].push_back(v);
	}
	cin>>s>>t;
	--s,--t;
	vector<int> d(n*3,inf);
	d[s]=0;
	queue<int> q;
	q.push(s);
	while(!q.empty())
	{
		int x=q.front();
		q.pop();
		int now=x%n;
		for(int i=0;i<graph[now].size();i++)
		{
			int dd=d[x]+1;
			int y=graph[now][i]+n*(dd%3);
			if(d[y]!=inf)
				continue;
			d[y]=dd;
			q.push(y);
		}
	}
	cout<<(d[t]==inf?-1:(d[t]/3))<<endl;
	return 0;
}
