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<int,int> P;
vector<int> G[100000];
ll dist[100000][3];
int main() {
	int N,M;
	cin>>N>>M;
	for(int i=0;i<M;i++) {
		int u,v;
		cin>>u>>v;u--;v--;
		G[u].push_back(v);
	}
	int S,T;
	cin>>S>>T;S--;T--;
	fill((ll*)dist,(ll*)(dist+N),1LL<<60);
	dist[S][0]=0;
	deque<P> Q;
	Q.emplace_back(S,0);
	while(!Q.empty()) {
		P now=Q.front();Q.pop_front();
		for(int to:G[now.first]) {
			ll cost = dist[now.first][now.second] + (now.second == 0);
			if(dist[to][(now.second+1)%3] > cost) {
				dist[to][(now.second+1)%3] = cost;
				if(now.second == 0) Q.emplace_back(to,1);
				else Q.emplace_front(to,(now.second+1)%3);
			}
		}
	}
	if(dist[T][0] == 1LL<<60) cout<<-1<<endl;
	else cout<<dist[T][0]<<endl;
}