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;
typedef long long ll;
typedef pair<ll,ll> P;
bool f[314514];
int main(){
	ll n,m;
	cin>>n>>m;
	vector<vector<ll>>v(314514);
	vector<ll>dis(314514,1e9);
	ll p=100000;
	for(int i=0;i<m;i++){
		ll a,b;
		cin>>a>>b;
		v[a].push_back(b+p);
		v[a+p].push_back(b+2*p);
		v[a+2*p].push_back(b);
	}
	ll s,t;
	cin>>s>>t;
	dis[s]=0;
	f[s]=1;
	queue<P>que;
	que.push({s,0});
	while(!que.empty()){
		P x=que.front();
		que.pop();
		for(ll num:v[x.first]){
			if(f[num])continue;
			f[num]=1;
			que.push({num,x.second+1});
			dis[num]=x.second+1;
		}
	}
	if(dis[t]==1e9)cout<<-1;
	else cout<<dis[t]/3;
	return 0;
}
