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<algorithm>
#include<cmath>
#include<cstring>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<deque>
#define ll long long
#define mp make_pair
using namespace std;
int n,m,bg,ed;
vector<int> v[100005];
int dp[100005][3];
inline void bfs(){
	memset(dp,63,sizeof(dp));
	dp[bg][0]=0;
	queue<pair<int,int> >q;
	pair<int,int> p;
	q.push(mp(bg,0));
	while(!q.empty()){
		p=q.front();
		q.pop();
		int u=p.first,x=p.second,y=(x+1)%3;
		if(u==ed and x%3==0){
			cout<<x/3<<endl;
			return;
		}
		for(int i=0;i<v[u].size();++i){
			int t=v[u][i];
			if(dp[t][y]>x+1){
				dp[t][y]=x+1;
				q.push(mp(t,x+1));
			}
		}
	}
	cout<<-1<<endl;
}
int main(){
	int i,j,x,y;
	scanf("%d%d",&n,&m);
	for(i=0;i<m;++i) scanf("%d%d",&x,&y),v[x].push_back(y);
	scanf("%d%d",&bg,&ed);
	bfs();
}