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>
using namespace std;
#define boost ios::sync_with_stdio(false);cin.tie(nullptr)
#define pi pair<int,int>
#define ff first
#define ss second
#define int long long
	

int32_t main(){
	boost;
	int n,m;
	cin>>n>>m;
	vector<int>g[n+1];
	for(int i=0,u,v;i<m;i++){
		cin>>u>>v;
		g[u].push_back(v);
	}
	int s,d;
	cin>>s>>d;
	queue<pi>q;
	q.push({s,0});
	int v[3][n+1];
	memset(v,0,sizeof v);
	v[0][s] = 1;
	while(!q.empty()) {
		pi u = q.front();
		q.pop();
		if(u.ff == d && u.ss%3 == 0){
			cout<<u.ss/3<<endl;
			return 0;
		}
		int x = u.ss + 1;
		for(int t:g[u.ff]){
			if(v[x%3][t] == 0){
				v[x%3][t] = 1;
				q.push({t,x});
			}
		}
	}
	cout<<-1<<endl;
}

