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 <vector>
#include <string.h>
#include <set>
#include <queue>
using namespace std;

vector<int> g[100003];
int memo[4][100003];

struct E{
	int p,c;
};

int main() {
	// your code goes here
	int n,m;
	cin>>n>>m;
	memset(memo,0,sizeof(memo));
	for(int i=0;i<m;i++){
		int u,r;
		cin>>u>>r;
		g[u].push_back(r);
	}
	int s,t;
	cin>>s>>t;
	
	

	int ans=-1;
	queue<E> qu;
	E e1;
	e1.p=s;
	e1.c=0;
	qu.push(e1);
	
	while (qu.empty()==false){
		E e1=qu.front();
		qu.pop();
		//cout<<"\n"<<e1.p<<" "<<e1.c<<")";
		if(e1.c%3==0&&e1.p==t){
			ans=e1.c/3;
			break;
		}
		
		if(memo[e1.c%3][e1.p]!=0){
			continue;
		}
		
		memo[e1.c%3][e1.p]=1;
		
		for(int i=0;i<g[e1.p].size();i++){
			E e2;
			e2.p=g[e1.p][i];
			e2.c=e1.c+1;
			qu.push(e2);
		}
	}
	cout<<ans<<"\n";
	return 0;
}