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<queue>
#include<vector>
#include<algorithm>
using namespace std;
int N,M;
vector<int>G[1<<17];
int S,T;
int d[1<<17][3];
int main()
{
	cin>>N>>M;
	for(int i=0;i<M;i++)
	{
		int a,b;cin>>a>>b;a--,b--;
		G[a].push_back(b);
	}
	for(int i=0;i<N;i++)for(int j=0;j<3;j++)d[i][j]=2e9;
	cin>>S>>T;S--,T--;
	queue<pair<int,int> >P;
	P.push({S,0});
	d[S][0]=0;
	while(!P.empty())
	{
		int u=P.front().first;
		int c=P.front().second;
		P.pop();
		for(int v:G[u])
		{
			if(d[v][(c+1)%3]>d[u][c]+1)
			{
				d[v][(c+1)%3]=d[u][c]+1;
				P.push({v,(c+1)%3});
			}
		}
	}
	cout<<(d[T][0]%3==0?d[T][0]/3:-1)<<endl;
}
