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 LL long long
#define M 100005
vector<int>E[M];
int n,m,s,t,a,b;
int Min=1e9;
struct Node {
	int x,y;
};
bool Mark[M][3];
queue<Node>Q;
void bfs(int x) {
	Q.push((Node)<%x,0%>);
	Mark[x][0]=1;
	while(!Q.empty()) {
		int x=Q.front().x,y=Q.front().y;
		Q.pop();
		if(x==t&&y%3==0)Min=min(Min,y);
		for(int i=0; i<E[x].size(); i++) {
			int u=E[x][i],w=(y+1)%3;
			if(!Mark[u][w])Q.push((Node)<%u,y+1%>),Mark[u][w]=1;
		}
	}
}
int main() {
	scanf("%d%d",&n,&m);
	for(int i=1; i<=m; i++) {
		scanf("%d%d",&a,&b);
		E[a].push_back(b);
	}
	scanf("%d%d",&s,&t);
	bfs(s);
	if(!Mark[t][0])puts("-1");
	else printf("%d",Min/3);
}