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>
using namespace std;
const int maxn = 123456;
vector<int> e[maxn];
int dist[4][maxn];
int main(){
	int n, m; cin >> n >> m;
	for(int i = 0; i < m; i++) {
		int x, y; cin >> x >> y;
		e[x].push_back(y);
	}
	memset(dist, -1, sizeof(dist));
	int s, t; cin >> s >> t;
	dist[2][s] = 0;
	deque<pair<int,int> > q; q.push_back({2, s});
	while(!q.empty()){
		int st, x; tie(st, x) = q.front(); q.pop_front();
		int nst = (st + 1) % 3;
		for(int u : e[x]){
			if(dist[nst][u] == -1){
				if(st == 2){
					dist[nst][u] = dist[st][x] + 1;
					q.push_back({nst, u});
				} else {
					dist[nst][u] = dist[st][x];
					q.push_front({nst, u});
				}
			}
		}
	}
	cout << dist[2][t] << endl;
	return 0;
}

