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 <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
using ll = long long;

int n, m, s, t;
vector<int> e[100010];
int dp[100010][3];
queue<int> q;
void bfs() {
	dp[s][1] = 1;

	q.emplace(s*3+1);
	while (!q.empty()) {
		int v = q.front() / 3, mod = q.front() % 3;
		q.pop();

		int nc = dp[v][mod] + 1;
		for (int d : e[v])
			if (!dp[d][nc % 3])
				dp[d][nc % 3] = nc, q.emplace(d*3+(nc%3));
	}
}
int main() {
	scanf("%d %d", &n, &m);
	int u, v;
	for (int i = 0; i < m; i++) {
		scanf("%d %d", &u, &v);
		e[u].emplace_back(v);
	}
	scanf("%d %d", &s, &t);
	bfs();
	if (dp[t][1])
		printf("%d\n", (dp[t][1] - 1) / 3);
	else
		printf("-1\n");
}