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;

const int INF = 1001001001;
vector<vector<int>> dist(100005, vector<int>(3, INF));

int main() {
	int n, m;
	cin >> n >> m;

	// 有向グラフG
	vector<vector<int>> to(n);
	for (int i = 0; i < m; i++)
	{
		int a, b;
		cin >> a >> b;
		a--; b--;
		to[a].push_back(b);
	}

	int sv, tv;
	cin >> sv >> tv;
	sv--; tv--;

	// BFS（幅優先探索）
	queue<pair<int, int>> q;
	q.push(pair<int, int>(sv, 0));
	dist[sv][0] = 0;
	while (!q.empty()) {
		int v = q.front().first;
		int l = q.front().second;
		q.pop();
		for (int u : to[v]) {
			int nl = (l + 1) % 3;
			if (dist[u][nl] != INF) continue;
			dist[u][nl] = dist[v][l] + 1;
			q.push(pair<int, int>(u, nl));
		}
	}
	int ans = dist[tv][0];
	if (ans == INF) ans = -1;
	else ans /= 3;
	cout << ans << endl;
	return 0;
}
