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;
const int SZ=300005;
vector<int> graph[SZ];
queue<int> q;
int dist[SZ];
int main(void){
	ios::sync_with_stdio(false); cin.tie(NULL);
	memset(dist,-1,sizeof(dist)); 
	int n,m; cin >> n >> m;
	for(int i=0;i<m;i++){
		int x,y; cin >> x >> y;
		graph[3*x].push_back(3*y+1);
		graph[3*x+1].push_back(3*y+2);
		graph[3*x+2].push_back(3*y);
	}
	int s,t; cin >> s >> t;
	dist[3*s]=0;
	q.push(3*s);
	while(!q.empty()){
		int f=q.front(); q.pop();
		for(int nxt:graph[f]){
			if(dist[nxt]!=-1) continue;
			q.push(nxt);
			dist[nxt]=dist[f]+1;
		}
	}
	if(dist[3*t]==-1) cout<<-1;
	else cout<<dist[3*t]/3;
	return 0;
}