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 ((int)1e5+5)
using namespace std;
int main(){
	// interesting
	int n,m;
	cin>>n>>m;
	vector<vector<int>> gra(N);
	for(int i=0;i<m;i++){
		int u,v;
		cin>>u>>v;
		u-=1;
		v-=1;
		gra[u].emplace_back(v);
	}
	int s,t;
	cin>>s>>t;
	s-=1;
	t-=1;
	queue<pair<int,int>> q;
	q.push(make_pair(0,s));
	vector<int> flag(3*n,-3);
	int dfsParent[3*n];
	memset(dfsParent,0,sizeof(dfsParent));
	while(!q.empty()){
		pair<int,int> nx=q.front();q.pop();
		int nxdis=nx.first;
		int nxpoi=nx.second;
		if(flag[nxpoi]!=-3)continue;
		flag[nxpoi]=nxdis;
		for(int i=0;i<(int)gra[nxpoi%n].size();i++){
			int nx=gra[nxpoi%n][i];
			q.push(make_pair(nxdis+1,(nxdis+1)%3*n+nx));
			dfsParent[(nxdis+1)%3*n+nx]=nxpoi;
		}
	}
	cout<<flag[t]/3<<endl;
	
	return 0;
}