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 X first
#define Y second
using namespace std;
typedef long long ll;

int n, m, a, b, p1, p2;
vector<int> v[100002];
int bio[100002][5];
queue< pair<int, int> > q;

void bfs() {
	
	while (!q.empty()) {
		int x = q.front().X;
		int y = q.front().Y;
		q.pop();
		for (int i = 0; i < v[x].size(); i ++) {
			int sus = v[x][i];
			if (bio[sus][(y+1) % 3] == -1) {
				bio[sus][(y+1) % 3] = bio[x][y]+1;
				q.push({sus, (y+1)%3});
			}
		}
	}

}

int main () {

	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> m;
	for (int i = 0; i < m; i++) {
		cin >> a >> b;
		v[a].push_back(b);
	}
	
	cin >> p1 >> p2;
	memset(bio, -1, sizeof bio);
	bio[p1][0] = 0;
	q.push({p1, 0});
	bfs();
	if (bio[p2][0] == -1) cout << -1;
	else cout << bio[p2][0]/3;

	return 0;
}