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<cstdio>
#include<queue>
#include<vector>
using namespace std;

const int INF = 123456789;
const int N_MAX = 100000;

vector<int> G[N_MAX];
queue<int> q;
int d[3*N_MAX];

int main() {
	int n, m, s, t;
	scanf("%d %d", &n, &m);
	for(int i=0;i<m;i++) {
		int u, v;
		scanf("%d %d", &u, &v);
		G[u-1].push_back(v-1);
	}
	scanf("%d %d", &s, &t);
	s--; t--;
	for(int i=0;i<3*n;i++) d[i] = INF;
	q.push(s);
	d[s] = 0;
	while(!q.empty()) {
		int v = q.front();
		if(v == t) break;
		int mod = v / n;
		int nxmod = (mod + 1) % 3;
		v %= n;
		q.pop();
		for(int i=0;i<G[v].size();i++) {
			if(d[mod*n+v] + 1 < d[nxmod*n+G[v][i]]) {
				d[nxmod*n+G[v][i]] = d[mod*n+v] + 1;
				q.push(nxmod*n+G[v][i]);
			}
		}
	}
	printf("%d\n", d[t] < INF ? d[t]/3 : -1);
	return 0;
}
