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>
#define For(a,b,c) for(int a=b;a<=c;++a)
#define Dor(a,b,c) for(int a=b;a>=c;--a)
using namespace std;
const int N=100007;
int n,m,s,e,hd[N],nt[N],to[N],Q[N*3],D[N*3],F[N*3],B[N][3];
int main() {
	scanf("%d%d",&n,&m);
	For(i,1,m) {
		int x,y;
		scanf("%d%d",&x,&y);
		nt[i]=hd[x],hd[x]=i,to[i]=y;
	}
	scanf("%d%d",&s,&e);
	Q[1]=s,D[1]=0;
	int h=0,t=1;
	while(h<t) {
		int u=Q[++h],d=D[h];
		for(int i=hd[u];i;i=nt[i])
		if(!B[to[i]][(d+1)%3]) {
			Q[++t]=to[i],D[t]=(d+1)%3,F[t]=F[h]+1,B[to[i]][(d+1)%3]=1;
			if(Q[t]==e&&!D[t]) {
				printf("%d\n",F[t]/3);
				return 0;
			}
		}
	}
	puts("-1");
	return 0;
}