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;

int n,m,u,v,s,tt,i,j,used[300005]={0},co[300005]={0};
vector<int> d[300005];
queue<int> que;

int main() {
	// your code goes here
	cin >> n >> m;
	for(i=1;i<=m;i++){
		cin >> u >> v;
		d[u].push_back(n+v);
		d[n+u].push_back(2*n+v);
		d[2*n+u].push_back(v);
	}
	for(i=1;i<=3*n;i++){
			co[i]=100000000;
	}
	cin >> s >> tt;
	co[s]=0;
	used[s]=1;
	que.push(s);
	while(!que.empty()){
		u=que.front();
		que.pop();
		for(i=0;i<d[u].size();i++){
			used[d[u][i]]=1;
			if(co[d[u][i]]>co[u]+1){
				co[d[u][i]]=co[u]+1;
				que.push(d[u][i]);
			}
		}
	}
	if(used[tt]==0) cout << "-1" << endl;
	else if(co[tt]==100000000) cout << "-1" << endl;
	else cout << co[tt]/3 << endl;
	return 0;
}