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 <iostream>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#define N 100005
using namespace std;

int n, m, s, t, d[N][3];
vector<int> gr[N];
queue<pair<int,int>> qu;

int main() {
	int i, t1, t2, u, c;
	cin >> n >> m;
	while(m--) {
		scanf("%d%d", &t1, &t2);
		gr[t1].push_back(t2);
	}
	memset(d, 0x7f, sizeof(d));
	cin >> s >> t;
	qu.push({s, 0});
	d[s][0] = 0;
	while(!qu.empty()) {
		u = qu.front().first;
		c = qu.front().second;
		qu.pop();
		for(int v:gr[u]) {
			if(d[u][c] + 1 < d[v][(c+1)%3]) {
				d[v][(c+1)%3] = d[u][c] + 1;
				qu.push({v, (c+1)%3});
			}
		}
	}
	cout << (d[t][0]%3 ? -1 : d[t][0]/3);
	return 0;
}