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 <iostream>
#include <vector>
#include <queue>
using namespace std;
const int MAX = 100000;
const int INF = 1001001001;

int main() {
	int n, m;
	cin >> n >> m;
	function<int(int, int)> idx = [&](int x, int m) {
		return m % 3 * n + x;
	};
	vector<vector<int>> g(n*3);
	for (int i = 0; i < m; i++) {
		int u, v;
		cin >> u >> v;
		u--; v--;
		for (int j = 0; j < 3; j++) {
			g[idx(u, j)].push_back(idx(v, j+1));
		}
	}
	int s, t;
	cin >> s >> t;
	s--; t--;
	int d[MAX*3]; fill(d, d+MAX*3, INF);
	d[s] = 0;
	queue<int> que;
	que.push(s);
	while (!que.empty()) {
		int v = que.front(); que.pop();
		for (int i : g[v]) {
			if (d[i] != INF) continue;
			d[i] = d[v] + 1;
			que.push(i);
		}
	}

	int ans = d[t]!=INF ? d[t] / 3 : -1;
	cout << ans << endl;
	return 0;
}