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;
#define rep(i,n) for(int i = 0; i < n; i++)
using P = pair<int, int>;
const int mx = 100005;
int n, m, u, v, s, t, dis[mx*3];
queue<int> qu;
vector<int> node[mx*3];
int main(){
	scanf("%d%d", &n, &m);
	rep(i,m){
		scanf("%d%d", &u, &v);u--;v--;
		node[u].push_back(v+mx);
		node[u+mx].push_back(v+mx*2);
		node[u+mx*2].push_back(v);
	}
	scanf("%d%d", &s, &t);s--;t--;
	qu.push(s); dis[s] = 1;
	while(!qu.empty()){
		int pos = qu.front(); qu.pop();
		for(int it : node[pos]){
			if(dis[it] == 0){
				qu.push(it);
				dis[it] = dis[pos]+1;
			}
		}
	}
	printf("%d\n", dis[t] ? dis[t]/3 : -1);
}