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 N 100005
#define debug(x) cout<<#x<<":"<<x<<endl;
using namespace std;
int n,m,s,t,ans;
vector<int>G[N];
bool mark[N][3];
bool flag=0;
struct node {
	int now,t,mod;
};
void bfs() {
	queue<node>q;
	q.push((node)<%s,0,0%>);
	mark[s][0]=1;
	while(q.size()) {
		node x=q.front();
		q.pop();
		for(int i=0; i<G[x.now].size(); i++) {
			int y=G[x.now][i],z=(x.mod+1)%3;
			if(mark[y][z])continue;
			mark[y][z]=1;
			if(y==t&&z==0) {
				flag=1;
				ans=x.t+1;
				ans/=3;
				return;
			}
			q.push((node)<%y,x.t+1,z%>);
		}
	}
}
int main() {
//	freopen("temp.txt","r",stdin);
	scanf("%d%d",&n,&m);
	for(int i=0; i<m; i++) {
		int x,y;
		scanf("%d%d",&x,&y);
		G[x].push_back(y);
	}
	scanf("%d%d",&s,&t);
	bfs();
	if(flag)printf("%d",ans);
	else puts("-1");
	return 0;
}
