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 MX = 1e5+2;
vector<int> adj[MX];
int dist[MX][3];
struct info {
	int u, dep;
};
int main() {
	ios::sync_with_stdio(0);cin.tie(0);
	int n, m;
	cin >> n >> m;
	while(m--) {
		int u, v;
		cin >> u >> v;
		adj[u].push_back(v);
	}
	int s, t;
	cin >> s >> t;
	memset(dist, 0x3f, sizeof dist);
	dist[s][0] = 0;
	queue<int> que;
	que.push(s);
	int d = 0, nxtd;
	while(que.size()) {
		int sz = que.size();
		nxtd = d == 2 ? 0 : d+1;
		while(sz--) {
			int q = que.front();
			que.pop();
			for(int a: adj[q]) {
				if(dist[q][d]+1 < dist[a][nxtd]) {
					dist[a][nxtd] = dist[q][d]+1;
					que.push(a);
				}
			}
		}
		d = nxtd;
	}
	cout << (dist[t][0] == 0x3f3f3f3f ? -1 : dist[t][0]/3);
}