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;

typedef long long ll;
typedef pair<int, int> pii;
const int N = 1e5 + 100, M = 3, inF = 1e9 + 100;

int n, m, s, t, dis[N][M];
vector<int> nei[N];
queue<pii> q;

void bfs(pii p) {
	int v = p.first, t = p.second;
	for (int u: nei[v]) {
		int nxt_v = u, nxt_t = (t + 1) % 3;
		if(dis[nxt_v][nxt_t] == inF) {
			dis[nxt_v][nxt_t] = dis[v][t] + 1;
			q.push(pii(nxt_v, nxt_t));
		}
	}
}

int main() {
	ios::sync_with_stdio(false), cin.tie(0);
	for (int i = 0; i < N; i++)
		fill(dis[i], dis[i] + M, inF);
	cin >> n >> m;
	while(m--) {
		int u, v;
		cin >> u >> v;
		nei[--u].push_back(--v);
	}
	cin >> s >> t;
	s--;
	t--;
	dis[s][0] = 0;
	q.push(pii(s, 0));
	while(q.size()) {
		bfs(q.front());
		q.pop();
	}
	cout << (dis[t][0] == inF? -1: dis[t][0] / 3);
	return 0;
}