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>
#define fi first
#define se second
#define mp make_pair
using namespace std;

typedef pair <int, int> pii;
typedef pair <pii, int> piii;

vector <int> from[100010];
int vis[100010][3];
queue <piii> q;

int main(){
	int n, m; scanf("%d%d", &n, &m);
	for (int i = 0; i < m; i++){
		int u, v; scanf("%d%d", &u, &v);
		from[u].push_back(v);
	}
	int S, T; scanf("%d%d", &S, &T);
	memset(vis, -1, sizeof(vis));
	q.push(mp(mp(S, 0), 0)), vis[S][0] = 0;
	while (!q.empty()){
		piii t = q.front(); q.pop();
		int x = t.fi.fi, p = (t.fi.se + 1) % 3, dist = t.se + 1;
		for (int i = 0; i < from[x].size(); i++){
			int v = from[x][i];
			if (~vis[v][p]) continue;
			vis[v][p] = dist, q.push(mp(mp(v, p), dist));
		}
	}
	printf("%d\n", vis[T][0] == -1 ? -1 : vis[T][0] / 3);
	return 0;
}
