#include <queue>
#include <vector>
#include <iostream>
using namespace std;
int main() {
	int N, M, S, T;
	cin >> N >> M;
	vector<vector<int> > G(N);
	for (int i = 0; i < M; ++i) {
		int a, b;
		cin >> a >> b; --a, --b;
		G[a].push_back(b);
	}
	cin >> S >> T; --S, --T;
	vector<vector<int> > dist(N, vector<int>(3, -1));
	dist[S][0] = 0;
	queue<pair<int, int> > que; que.push(make_pair(S, 0));
	while (!que.empty()) {
		pair<int, int> u = que.front(); que.pop();
		for (int i : G[u.first]) {
			int nxt = (u.second + 1) % 3;
			if (dist[i][nxt] == -1) {
				dist[i][nxt] = dist[u.first][u.second] + 1;
				que.push(make_pair(i, nxt));
			}
		}
	}
	if (dist[T][0] == -1) cout << -1 << endl;
	else cout << dist[T][0] / 3 << endl;
	return 0;
}
### Prompt

In CPP, your task is to solve the following problem:

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

### Response

```cpp
#include <queue>
#include <vector>
#include <iostream>
using namespace std;
int main() {
	int N, M, S, T;
	cin >> N >> M;
	vector<vector<int> > G(N);
	for (int i = 0; i < M; ++i) {
		int a, b;
		cin >> a >> b; --a, --b;
		G[a].push_back(b);
	}
	cin >> S >> T; --S, --T;
	vector<vector<int> > dist(N, vector<int>(3, -1));
	dist[S][0] = 0;
	queue<pair<int, int> > que; que.push(make_pair(S, 0));
	while (!que.empty()) {
		pair<int, int> u = que.front(); que.pop();
		for (int i : G[u.first]) {
			int nxt = (u.second + 1) % 3;
			if (dist[i][nxt] == -1) {
				dist[i][nxt] = dist[u.first][u.second] + 1;
				que.push(make_pair(i, nxt));
			}
		}
	}
	if (dist[T][0] == -1) cout << -1 << endl;
	else cout << dist[T][0] / 3 << endl;
	return 0;
}
```