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"

typedef long long ll;

const int size = 100005;

std::vector<int> g[size];

bool visited[size][3];

int dfs(int s, int e)
{
	// Vertex, mod
	std::queue<std::pair<int, int>> q;
	q.push({ s, 0 });
	while (!q.empty())
	{
		int vert = q.front().first;
		int ken = q.front().second;
		q.pop();
		if (vert == e && ken % 3 == 0)
		{
			return ken / 3;
		}
		if (visited[vert][ken % 3])
		{
			continue;
		}
		visited[vert][ken % 3] = true;
		for (int nV : g[vert])
		{
			q.push({ nV, ken + 1 });
		}
	}
	return -1;
}

int main()
{
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i = 0; i < m; ++i)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		g[u].push_back(v);
	}
	int s, e;
	scanf("%d%d", &s, &e);
	printf("%d\n", dfs(s, e));
	return 0;
}