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;
#define int long long
struct vl {
	int u, step;
};
const int MAXN = 1e5 + 69;
int n, m;
int s, t;
vector<int> a[MAXN];
int check[MAXN][3];
void bfs(vl u) {
	queue<vl> q;
	check[u.u][u.step] = 1;
	q.push(u);
	while (!q.empty()) {
		vl u = q.front(); q.pop();
		for (int v : a[u.u]) {
			int nstep = (u.step + 1) % 3;
			if (check[v][nstep]) continue;
			check[v][nstep] = check[u.u][u.step] + 1;
			q.push({v, nstep});
		}
	}
}
signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	cin >> n >> m;
	for (int i = 1, u, v; i <= m; i++) {
		cin >> u >> v;
		a[u].push_back(v);
	}
	cin >> s >> t;
	bfs({s, 0});
	if (!check[t][0]) cout << -1;
	else cout << (check[t][0] - 1) / 3;
}