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;
#define rep(i,n) for(int i=0;i<(n);i++)


vector<int> g[300010];
bool used[300010];

int main(int argc, char const *argv[])
{
	int n, m; cin >> n >> m;
	vector<int> u(m), v(m); rep(i, m) cin >> u[i] >> v[i];
	rep(i, m) u[i]--, v[i]--;
	int s, t; cin >> s >> t;
	s--, t--;

	rep(i, m) {
		rep(j, 3)
			g[u[i] * 3 + j].push_back(v[i] * 3 + ((j + 1) % 3));
	}

	queue<pair<int, int>> q;
	q.push(make_pair(3 * s, 0));
	used[3 * s] = true;
	while (!q.empty()) {
		int d = q.front().first, cost = q.front().second;
		q.pop();

		if (d == 3 * t) {
			printf("%d\n", cost / 3);
			return 0;
		}
		for (auto nd : g[d]) {
			if (!used[nd]) {
				used[nd] = true;
				q.push(make_pair(nd, cost + 1));
			}
		}
	}
	printf("-1\n");
	return 0;
}
