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 vi vector<int>
#define pb push_back
using namespace std;
const int N=1e5+5;
bool d[3][N];
int s,t,n,m,dist[3][N];
vi v[N],e1,e2;//可以制造余数1和余数2的 
queue<int>q;
int main()
{
	for(int i=0;i<N;i++)dist[0][i]=dist[1][i]=dist[2][i]=1e9;
	cin>>n>>m;
	for(int i=0;i<m;i++)
	{
		int a,b;
		cin>>a>>b;
		a--,b--;
		v[a].pb(b);
	}
	cin>>s>>t;
	s--,t--;
	q.push(s);
	q.push(0);
	while(!q.empty())
	{
		int now=q.front();q.pop();
		int move=q.front();q.pop();
		if(d[move%3][now])continue;
		dist[move%3][now]=move;
		d[move%3][now]=true;
		for(int i=0;i<v[now].size();i++)
		{
			int to=v[now][i];
			q.push(to);
			q.push(move+1);
		}
	}
	if(d[0][t])
	{
		cout<<dist[0][t]/3<<endl;
	}
	else
	{
		cout<<-1<<endl;
	}
	return 0;
}