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;
const int N=1e5+5;
int n,m,h[N],w[N],g[N],x,y,s,t;
struct E{
	int l,to;
}e[N];
struct P{
	int d,w;
};
queue<P>Q;
int main(){
	scanf("%d %d",&n,&m);
	for(int i=1;i<=m;++i){
		scanf("%d %d",&x,&y);
		e[i].to=y;
		e[i].l=h[x];
		h[x]=i;
	}
	scanf("%d %d",&s,&t);
	w[s]=1;
	Q.push({s,0});
	while(!Q.empty()){
		P k=Q.front();Q.pop();
		bool fl= (k.w%3)==2;
//		printf("%d %d %d\n",k.d,k.w,fl);
		for(int i=h[k.d];i;i=e[i].l){
			if(fl){
				if(!w[e[i].to]){
					w[e[i].to]=k.w/3+1;
					Q.push({e[i].to,k.w+1});
				}
			}
			else {
				if(g[e[i].to]!=k.w+1){
					g[e[i].to]=k.w+1;
					Q.push({e[i].to,k.w+1});
				}
			}
		}
		if(w[t])break;
	}
	if(!w[t])w[t]=-1;
	printf("%d",w[t]);
}