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;
#define M 100001
struct D{
	int x,step;
}nxt;
int n,m,st,ed;
vector<int>edge[M];
bool mark[M][3];
int spfa(){
	queue<D>Q;
	Q.push((D){st,0});mark[st][0]=1;
	while(!Q.empty()){
		D now=Q.front();Q.pop();
		for(int i=0;i<edge[now.x].size();i++){
			int y=edge[now.x][i];
			nxt.x=y,nxt.step=now.step+1;
			if(mark[y][nxt.step%3])continue;
			mark[y][nxt.step%3]=1;
			if(y==ed&&nxt.step%3==0)return nxt.step/3;
			Q.push(nxt);
		}
	}
	return -1;
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1,u,v;i<=m;i++){
		scanf("%d%d",&u,&v);
		edge[u].push_back(v);
	}
	scanf("%d%d",&st,&ed);
	printf("%d",spfa());
	return 0;
}