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 sz(c)      int(c.size())
#define rep(i,a,b) for (int i=a; i<(b); ++i)
#define per(i,a,b) for (int i=(b)-1; i>=(a); --i)
using namespace std;
using ll = long long;

int const MAXN=110000;
int N,M,S,T;
vector<int> adj[MAXN];
int d[MAXN][3];

int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(0);
	cout<<fixed<<setprecision(10);

	cin>>N>>M;
	rep(i,0,M) {
		int x,y;
		cin>>x>>y;
		x--,y--;
		adj[x].push_back(y);
	}
	cin>>S>>T;
	S--,T--;
	
	rep(i,0,N) rep(j,0,3) d[i][j]=-3;
	
	d[S][0]=0;
	queue<pair<int,int>> q;
	q.emplace(S,0);
	while (!q.empty()) {
		int v,k;
		tie(v,k)=q.front();
		q.pop();
		
		int nk=(k+1)%3;
		for (int u:adj[v]) if (d[u][nk]==-3) {
			d[u][nk]=d[v][k]+1;
			q.emplace(u,nk);
		}
	}
	
	cout<<d[T][0]/3<<"\n";
}
