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;

int n, m, s, t;
vector<int>adjl[100002];
int dist[100002][3]; 
queue<pair<int,int> >bfs;

int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);
	memset(dist,-1,sizeof dist);
	cin >> n >> m;
	for (int i=0;i<m;i++){
		int u, v;
		cin >> u >> v;
		adjl[u].push_back(v);
	}
	cin >> s >> t;
	bfs.push({s,0});
	while (!bfs.empty()){
		int now=bfs.front().first;
		int distnow=bfs.front().second;
		bfs.pop();
		if (dist[now][distnow%3]!=-1) continue;
		dist[now][distnow%3]=distnow;
		for (int next : adjl[now]) bfs.push({next,distnow+1});
	}
	cout << (dist[t][0]==-1 ? -1 : dist[t][0]/3)  << "\n";
}