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;



ll N, M;

vector<int> next_node[100000];

int main(){
	cin >> N >> M;
	
	for ( int i = 0; i < M; i++ ){
		int u, v;
		cin >> u >> v;
		u--;
		v--;
		next_node[u].push_back(v);
	}
	
	int S, T;
	cin >> S >> T;
	S--;
	T--;
	
	queue<pair<int,int>> q;
	map<int,bool> visit[3]; // 到達予定なら、印をつける
	
	q.push({S,0});
	while ( !q.empty() ){
		auto a = q.front();
		q.pop();
		int node = a.first;
		int step = a.second;
		
		if ( step % 3 == 0 ){
			if ( node == T ){
				cout << step/3 << endl;
				return 0;
			}
		}
		
		step++;
		for ( auto it : next_node[node] ){
			if ( visit[step%3][it] )
				continue;
			visit[step%3][it] = true;
			q.push({it,step});
		}
	}
	
	cout << -1 << endl;
	return 0;
}

