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 N = 1e5+5;
const int INF = 1e9;
int dist[N][3];
int main(){
	int n,m,s,t;
	cin>>n>>m;

	vector<int> G[n];
	for(int i = 0; i < n; i++){
		dist[i][0] = dist[i][1] = dist[i][2] = INF;
	}

	for(int i = 0; i < m; i++){
		int x,y;
		cin >> x >> y;
		--x,--y;
		G[x].push_back(y);
	}
	cin>>s>>t;
	--s,--t;
	dist[s][0] = 0;
	queue<pair<int,int>> q;
	q.push({s,0});
	while(!q.empty()){
		int u = q.front().first;
		int l = q.front().second;
		//cout<<u<<" "<<l<<endl;
		q.pop();
		for(int v : G[u]){
			 int nl = (l+1)%3;
			if(dist[v][nl]!=INF) continue;
			dist[v][nl]=dist[u][l]+1;
			q.push({v,nl});
		}
	}

	int ans = dist[t][0];
	if(ans == INF) cout<<-1;
	else cout<<(ans/3);	
}