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>
#define X first
#define Y second
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;

const ll Q = (ll)(1e9) + 7;
const int MAXN = 1e5+1;

vector<int> nxt[MAXN];
vector<pii> q;
bool used[3][MAXN];
int dist[3][MAXN];



int main() {
	ios_base::sync_with_stdio(0);

	int n, m;
	cin >> n >> m;
	while(m--) {
		int x, y;
		cin >> x >> y;
		nxt[x].push_back(y);
	}
	int s, t;
	cin >> s >> t;
	q.push_back({s,0});
	used[0][s] = 1;
	for(int i = 0; i < q.size(); i++) {
		int v = q[i].X;
		int d = q[i].Y;
		for(int w : nxt[v]) {
			int dd = (d+1)%3;
			if(!used[dd][w]) {
				used[dd][w] = 1;
				dist[dd][w] = dist[d][v] + 1;
				q.push_back({w,dd});
			}
		}
	}

	cout << (used[0][t] ? dist[0][t]/3 : -1) << '\n';

	return 0;
}