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<iostream> 
#include<algorithm>
#include <cstdio>
#include <cmath>
#include <vector>
#include <queue>
#include <cstring> 
using namespace std;
const int N=1e5+10;
const int INF=1e9;
int n,m,s,t,dis[N][3];
vector<int> G[N];
void bfs(){
	memset(dis,-1,sizeof(dis));
	dis[s][0]=0;
	queue<pair<int,int> >que;
	que.push({s,0});
	while(!que.empty()){
		pair<int,int> u=que.front();que.pop();
		for(int i=0;i < G[u.first].size();i++){
			if(dis[G[u.first][i]][(u.second+1)%3]<0){
				dis[G[u.first][i]][(u.second+1)%3]=dis[u.first][u.second]+1;
				que.push({G[u.first][i],(u.second+1)%3});
			}
		}
	}
	
}
int main(){
	cin>>n>>m;
	int u,v;
	while(m--){
		cin>>u>>v;
		G[u].push_back(v);
	}
	cin>>s>>t;
	bfs();
	if(dis[t][0]==-1)puts("-1");
	else cout<<dis[t][0]/3<<endl;
	return 0;
}
