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,HEAD[100005],b,s,e,H[3][100005],Q[100005],h,t,q,C[100005];
struct V {
	int e,n;
} B[100005];
void ADD(int s,int e) {
	B[++b].n=HEAD[s];
	HEAD[s]=b,B[b].e=e;
}
int main() {
	scanf("%d %d",&n,&m);
	for(int i=1; i<=m; i++)
		scanf("%d %d",&s,&e),ADD(s,e);
	scanf("%d %d",&s,&e);
	H[0][s]=1;
	Q[++t]=s;
	while(t>h) {
		q=Q[++h];
		for(int z=HEAD[q],i; z; z=B[z].n) {
			i=B[z].e;
			if(!H[1][i]) {
				H[1][i]=1;
				for(int y=HEAD[i],j; y; y=B[y].n) {
					j=B[y].e;
					if(!H[2][j]) {
						H[2][j]=1;
						for(int x=HEAD[j]; x; x=B[x].n) {
							if(!H[0][B[x].e]) {
								Q[++t]=B[x].e,H[0][B[x].e]=1,C[B[x].e]=C[q]+1;
							}
						}
					}
				}
			}
		}
	}
	if(!H[0][e])printf("-1");
	else printf("%d",C[e]);
	return 0;
}